Bernoulli Numbers

The Bernoulli numbers were first identified by Jakob Bernoulli in Ars Conjectandi, a work that was published after his death. The Bernoulli numbers turn up again and again in the history of mathematics. In today's blog, I will focus on the original problem that Jakob Bernoulli was attempting to solve. In a future blog, I will show how they were used by Ernst Kummer as a criteria for determining if a prime is regular.

Interestingly, Bernoulli numbers may have been discovered earlier by Seki Kowa and were rediscovered independently by a 17-year-old Srinivasa Ramanujan.

Bernoulli was analyzing formulas for the sums of powers, that have the following form:

S_m(n) = ∑ (i=1, n) i^m

He looked at, for example:

S₀(n) = n

S₁(n) = (1/2)n² - (1/2)n

S₂(n) = (1/3)n³ - (1/2)n² + (1/6)n

S₃(n) = (1/4)n⁴ - (1/2)n³ + (1/4)n²

S₄(n) = (1/5)n⁵ - (1/2)n⁴ + (1/3)n³ - (1/30)n

S₅(n) = (1/6)n⁶ - (1/2)n⁵ + (5/12)n⁴ - (1/12)n²

S₆(n) = (1/7)n⁷ - (1/2)n⁶ + (1/2)n⁵ - (1/6)n³ + (1/42)n.

S₇(n) = (1/8)n⁸ - (1/2)n⁷ + (7/12)n⁶ - (7/24)n⁴ + (1/2)n²

S₈(n) = (1/9)n⁹ - (1/2)n⁸ + (2/3)n⁷ - (7/15)n⁵ + (2/9)n³ - (1/30)n

S₉(n) = (1/10)n¹⁰ - (1/2)n⁹ + (3/4)n⁸ - (7/10)n⁶ + (1/2)n⁴ - (3/20)n²

S₁₀(n) = (1/11)n¹¹ - (1/2)n¹⁰ + (5/6)n⁹ - n⁷ + n⁵ - (1/2)n³ + (5/66)n.

By looking at these formulas, he noticed a pattern and in analyzing this pattern, he came up with what were termed Bernoulli numbers by Abraham de Moivre.

Well, for each S_m(n), we can see that:

1) the first term is 1/(m+1)*n^m+1

2) the second term, when it exists, is (-1/2)n^m

3) the third term, when it exists, is (m/12)*n^m-1

4) the fourth term, when it exists, is (-m)(m-1)(m-2)/720*n^m-3

Bernoulli had discovered that if he defined a sequence of the numbers, he got a simple formula for defining the sum of powers.

The formula he discovered was the following:



In the above formula, B_k refers to the sequence of Bernoulli numbers where Bernoulli numbers are defined in the following way:

Definition 1: Bernoulli Numbers



with B₀ = 1.

As a convention, I will refer to Bernoulli numbers as B_i.

So, to figure out, B₁, we have:



This gives us:

2*B₀ + 2*B₁ = 0

B₁ = (1/2)[-2*B₀] = (-1/2)(1) = (-1/2)

We determine B₂ based on solving:



In this way, we find that:

B₀ = 1
B₁ = -(1/2)
B₂ = (1/6)
B₃ = 0
B₄ = -(1/30)
B₅ = 0
B₆ = (1/42)
B₇ = 0
B₈ = -(1/30)
B₉ = 0
B₁₀ = (5/66)
B₁₁ = 0
B₁₂ = -(691/2730)

Bernoulli wrote that he was able to use his formula to compute S₁₀(1000) in less than 10 minutes (see Smith, A Source Book in Mathematics, below).

Using his formula, this comes down to:





Evaluating this gives us:

91,409,924,241,424,243,424,241,924,242,500

I will now show a proof for Bernoulli's formula for sums. This is taken straight from Graham et al. Concrete Mathematics.

Definition 2: S_m(n)

Let S_m(n) = 0^m + 1^m + ... + (n-1)^m

I will use this shorthand in the lemma and theorem below.

Postulate 1: 0⁰ = 1

(This is taken from Graham et al.)

Graham et al argue in Concrete Mathematics that if we let 0⁰ = 1, it provides a more elegant notation for the Binomial Theorem. Their argument is that the Binomial Theorem is very important while 0⁰ is almost irrelevant. (see Concrete Mathematics, page 162).

Lemma 1:



Proof:

(1) S_m+1(n) + n^m+1 =

= ∑(k=0,n-1) k^m+1 + n^m+1 =

= ∑(k=0,n) k^m+1 =

= 0^m+1 + ∑(k=1,n) k^m+1 = ∑(k=1,n) k^m+1 =

= ∑ (k=0,n-1) (k+1)^m+1

(2) Using the Binomial Theorem, we get:



(3) Putting this together with step #1 gives us:



This gives us:



(4) Since S_j(n) = o^j + 1^j + ... + (n-1)^j, we have shown that:



(5) Now, since



The conclusion follows from simple subtraction since [(m+1)!]/[(m+1)!(0)!] = 1.

QED

Theorem:



Proof:

(1) We can see that this is true for m=0 since:

S₀ = n = (1/[1+1])∑ (k=0,0) ([0+1]!/k![1-k]!)B_kn^0+1-k =

= (1)*(1!/[0!][1!])B₀n¹ = (1)*(1)*(1)*n = n [Since B₀ = 1, see above]

(2) Let's assume that this is true up to m-1.

(3) Let T_m(n) =



(4) Let Δ = S_m(n) - T_m(n)

(5) Using Lemma 1 above, we know that:



(6) Using our definition of Δ, we can also see that:



(7) The goal in this proof will be to show that Δ = 0.

(8) Adding the definition of T_j(n) [see step#3] gives us:



I also simplified [(m+1)!/(m!)(1!)] to (m+1).

(9) We can move all values that are independent of k before the ∑ (k=0,j):



(10) Since we can change the order of the terms of a finite sum and still get the same result, we can switch each value of k with (j-k) to get:






(11) Now, since (j+1) - (j-k) = k+1, we have:



(12) Putting this result in step #11 gives us:



(13) We can see that in step #12, 0 ≤ k ≤ j ≤ m. The sum is in fact all combinations of j,k where this is the case. This means that we can rearrange the sum to the following:



(14) Since n^k+1 is independent of j, we can move it over to get the following:



(15) Since,



we can also rearrange the result in step #14 to give us:



(16) Since:











We can rearrange terms to get:



(17) Since:



We can rearrange terms to get:



(18) Now, we know from the definition of Bernoulli numbers (see Definition 1 above) that



in all cases except where m = 0. This occurs only when k = m.

This allows us to simplify the result in step #17 to the following:



(19) Now since [(m+1)!/(m!)(1!)] = m+1, we get:

= n^m+1 + (m+1)Δ

(20) But in step #5, we started out with n^m+1 so we are left with:

n^m+1 = n^m+1 + (m+1)Δ.

(21) The only way that this is true is if Δ = 0.

QED

In another blog, I show how z/[e^z - 1] can be used as a generating function for the Bernoulli numbers.

References

• Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, Addison-Wesley, 1989
• David Eugene Smith, A Source Book in Mathematics, Dover, 1959
• "Bernoulli Numbers", Wikipedia.Com
  

Johann Bernoulli

Johann Bernoulli was born July 27, 1667 in Basel Switzerland. He was the tenth born child of a wealthy, prominent Swiss family. His father tried to interest him in the thriving family spice business but it did not work out. When Johann was 15, he began working for his father's company but this only lasted a year.

Johann started at the University of Basel in 1683. His father encouraged him to study medicine. His older brother Jacob had already begun to make a name for himself in mathematics and physics. Johann soon took an interest to math and studied under his brother's guidance. For two years, the brothers studied the new methods of Gottfriend von Leibniz which had recently been published. Johann showed a tremendous ability for the new mathematics and after these two years, in many ways became the mathematical equal of his more experienced, older brother.

In 1690, Johann published his first scientific paper on the process of fermentation. Later, in 1691, he went to Geneva to lecture on the new differential calculus. He traveled to France, like his brother had, and met with the followers of Rene Descartes who were led by Nicolas Malebranche. In France, Johann met Gillaume Francois Antoine Marquis de L'Hopital who was considered one of the finest of the Paris mathematicians. Johann taught L'Hopital about the calculus and Leibniz's methods. L'Hopital paid Johann well for the lessons and Johann continued the lessons via correspondence after he returned to the University of Basel.

Even while Johann studied medicine, he was continually publishing mathematical papers. He even began a very substantial correspondence with Leibniz on the ideas of calculus. In 1694, when Johann submitted his paper on medicine, the content of the paper was the mathematics of muscular movement.

He continued to generate significant mathematical papers and in 1695, he was offered the chair of mathematics at Groningen. This was offered based on the recommendation of Christiaan Huygens. Johann was glad to accept even though the move to Groningen was difficult.

While in Groningen, he got involved in many controversies. One of his medical lectures was criticized as casting doubt on the possibility of ressurection. One of his students accused him of abandoning Calvinism in order to embrace the philosophy of Descartes. Johann and his brother Jacob also began a fierce rivalry as each brother tried to outdo the other.

L'Hopital would later publish the first book on calculus. This book contained the famous L'Hopital's Rule which may have come from Johann Bernoulli.

Johann became a bit bothered by L'Hopital's book because he did not feel that it paid him proper acknowledgment. L'Hopital wrote in the book (quoted on the MacTutor web site):

And then I am obliged to the gentlemen Bernoulli for their many bright ideas; particularly to the younger Mr Bernoulli who is now a professor in Groningen.

When L'Hopital died in 1704, Johann insisted that most of the content of the book had been taken from him. At the time, few believed him. In 1922 (according to the MacTutor web site), evidence surfaced which substantiated Johann's claims which included mathematical proofs that were virtually identical to the proofs included L'Hopital's book.

Johann had married shortly before moving to Gottigen and in 1705, he learned that his father-in-law was very sick and not expected to live much longer. Johann and his family moved back to Basel. Once there, he learned that his father-in-law had made a recovery but his brother Jacob had died of tuberculosis. Johann decided to stay in Basel and took over after his brother as the Chair of Mathematics at the University of Basel.

In 1713 when Leibniz and Sir Isaac Newton quarrelled about who came first with regard to calculus. Johann weighed in on the size of Leibniz. Johann argued that Leibniz's methods were more powerful than Newton's (a position that is generally supported today) and that Descarte's theory of vortex was superior to Newton's theory of gravitation. This controversy has been documented in a new book that came out this year.

Three of Johann's sons would go on to become famous mathematicians. In 1734, his son Daniel published what would become one of his most important books. Johann proceeded to publish his own version of the same topic and claimed to have finished it in 1732. It is interesting that such a giant in mathematics would feel the need to compete with his own son.

Johann died on January 1, 1748 in Basel, Switzerland. At this time, his fame was giant. His contributions had been recognized by the scientific academies of Paris, Berlin, London, St. Petersburg, and Bologna.

References

• Biography of Johann Bernoulli, MacTutor
  

Jacob Bernoulli

Jacob Bernoulli was the oldest of the famous Bernoulli mathematicians. His younger brother Johann and his nephew Daniel all went on to become very famous and important mathematicians.

Jacob was born on December 27, 1654 to a prominent and wealthy family of merchants in Basel, Switzerland. His grandfather had started a prosperous spice business in Amsterdam but then had immigrated to Switzerland when Spain began a crackdown on protestants in 1567.

Jacob's father, Nicholas, was a member of the town council and a magistrate. His mother was the daughter of an important Basel banker. When the time came for Jacob to enter the university, he was strongly encouraged by his parents to study philosphy and theology. This apparently was the pathway to becoming a minister. He graduated from the University of Basel in 1676 with a licentiate in theology and a master's degree in philosophy. This same year, he moved to Geneva where he worked as a tutor.

All throughout his coursework, Jacob had also been studying astronomy and mathematics. Despite his parents discouragement of these pursuits, Jacob later traveled to France to study with the followers of Descartes who were led by Nicholas Malenbranche. In 1681, Jacob traveled to the Netherlands where he met with numerous mathematicians including Johanne Hudde. After this, he traveled to England where he met with Robert Boyle, Robert Hooke, and many others. In these travels, Bernoulli made numerous contacts and developed an extensive correspondence with many of the most influential mathematicians and scientists of his day. He would be offered a position with the church in Switzerland, to the great joy of his parents, but he turned it down.

He returned to the University of Basel in 1683 to teach courses in mathematics and theoretical physics. Bernoulli became fascinated with infinitesimal geometry as well as the works of Franz van Schooten, John Wallis, Isaac Barrow, and of course, the Geometrie by Rene Descartes. Jacob published many of his ideas in the math journal Acta Eruditorium which was founded in Leipzig in 1682.

Around this time, Jacob married Judith Stupanis and had two children. Neither of whom became interested in mathematics or physics. This is interesting considering the large number of Bernoullis who would become famous scientists. The New Dictionary of Scientific Biography, for example, lists 13 Bernoullis in its biographies.

Jacob's younger brother Johann was encouraged by the family to study medicine. While pursuing a medical degree, Johann was taught mathematics by Jacob. Both brothers became especially interested in Gottfried Wilhelm von Leibniz's paper on differential calculus which had just been published in Acta Eruditorum in 1684. At that time, Leibniz's paper had not yet made the impact that it would later make so the Bernoulli's interest in it represented a deeper understanding of the ideas involved. As Johann's mathematical abilities began to shine bright, there arose a falling out between the two brothers which would later become an out-and-out rivalry. The MacTutor Biography of Jacob Bernoulli says:

"...both made contributions to mathematics of the very greatest importance. Whether the rivalry spurred them on to greater things or whether they might have achieved more had they continued their initial collaboration, it is impossible to say."

Jacob Bernoulli went on to make a very significant mark on the mathematical world. He wrote on a wide range of topics including the parallels between logic and algebra, probability, infinite series, geometry, and calculus. In his geometry, he offered a method for dividing a triangle into four equals parts with two parallel lines; in probability, he offered a mathematical interpretation of relative frequency; on infinite series, he wrote about ∑ 1/n² which later became known as the Basel Problem (and would later be solved by Leonhard Euler) and wrote about exponential series that would later lead Euler to his creation of the constant e. In calculus, he wrote about the isochrone (a curve of constant descent that had been studied by Christiaan Huygens and Lebiniz) and showed how its path can be characterized by a first order nonlinear differential equation (this paper is also historically the first time that the term "integral" is used in the modern sense).

Jacob's other contributions include his analysis of what is today known as the Bernoulli equation: y'= p(x)y + q(x)yⁿ, a general method for determining the evolutes of a curve, did important work with parabolas, logarithmic spirals, epicycloids, what is today known as the lemniscate of Bernoulli, and the Drawbridge Problem (a problem regarding the keeping of a weight and draw bridge balanced).

Jacob Bernoulli became the chair of mathematics and the University of Basel and held this post until his death in 1705. Eight years after his death, his greatest work Ars Conjectandi was published. It was unfinished but would have a great impact on the theory of probability and introduced the idea of what would later be called Bernoulli Numbers.

The Dictionary of Scientific Biography (as quoted from the MacTutor web site) wrote this about Jacob Bernoulli's contribution:

Bernoulli greatly advanced algebra, the infinitesimal calculus, the calculus of variations, mechanics, the theory of series, and the theory of probability. He was self-willed, obstinate, aggressive, vindictive, beset by feelings of inferiority, and yet firmly convinced of his own abilities. With these characteristics, he necessarily had to collide with his similarly disposed brother. He nevertheless exerted the most lasting influence on the latter.

Bernoulli was one of the most significant promoters of the formal methods of higher analysis. Astuteness and elegance are seldom found in his method of presentation and expression, but there is a maximum of integrity.

After his death, his brother Johann became the chair of mathematics at the University of Basel.

References

• Biography of Jacob Bernoulli, MacTutor Web Site
  

The odds of two integers being relatively prime

In today's blog, I will use the zeta function to answer the question: what are the odds that two integers selected at random are relatively prime.

Lemma 1: The odds that p divides any positive integer selected at random is 1/p

Proof:

(1) Let us assume that we have n integers, that is, the integers 1..n = 1, 2, 3, ..., n. Let x be an integer randomly selected from this list.

(2) If n ≡ 0 (mod p), then it is clear that the odds are 1/p since:

(a) Since p divides n, there exists a value n' such that n = pn'.

(b) So we can divide up n integers in the following way into n' groups of p:

1, ... , p
p+1, ..., 2p
...
n'p-1, ... n'p

(c) We can see that regardless of group n falls in, 1 out of p of those integers in each group are divisible by p.

(3) if n ≡ a (mod p) where a ≠ 0, then we know that p divides n-a so that we have (n-a)=pn'

(4) In this case, we divide up the n integers into n' groups of p just as before but then we are also left with a group of a integers.

1, ..., p
p+1, .. , 2p
...
n'p-1, ..., n'p
n'p+1, ... n'p+a

(5) We can see that there are a total of n' ways to be divisible by p out of a total of n integers. This gives us odds = n'/(n'p+a)

(6) As we increase n toward infinity, we can use L'Hopital's Rule to give us:

lim (n' → ∞) [n'/(n'p + a)]

let f(x) = n'
let g(x) = n'p + a

f'(x) = 1
g'(x) = p

So lim(n' → ∞) [n'/(n'p+a)] = f'(x)/g'(x) = 1/p

QED

Corollary 1.1: The odds that p divides any 2 positive integers selected at random is 1/p²

Proof:

This follows directly from Lemma 1 above. If the odds of p dividing each positive integer is 1/p, the odds of it dividing both are (1/p)(1/p) = 1/p²

QED

Lemma 2: The odds that any 2 positive integers selected at random are relatively prime is 6/π²

Proof:

(1) Let x,y be two positive integers selected at random.

(2) The odds that a prime p divides both is 1/p² [From the Corollary 1.1 above]

(3) The odds that a prime p fails to divide both is (1 - 1/p²)

(4) Since each prime is distinct, the odds that no prime p divides both = ∏ (p = all primes)(1 - 1/p²)

(5) Now, let's consider the reciprocal of each factor (1 - 1/p²):

1/(1 - 1/p²) = 1/([p² - 1]/p²) = p²/(p² - 1)

(6) Now, consider what happens if we multiply to (p² - 1) to the following sequence S:

Let S = 1 + 1/p² + 1/p⁴ + 1/p⁶ + 1/p⁸ + ...

(p² - 1)S = p² + 1 + 1/p⁴ + ... - 1 - 1/p² -1/p⁴ = p²

So that:

S = p²/(p² - 1) = 1 + 1/p² + 1/p⁴ + ...

(7) Putting this together gives us:

∏ (p = all primes)1/(1 - 1/p²)=∏ (p = all primes)(p²/(p² - 1) = ∏ (p = all primes) ∑ (n=1,∞) 1/p^2(n-1) = (1 + 1/2² + 1/2⁴ + ... )(1 + 1/3² + 1/3⁴ + ...)(1 + 1/5² + 1/5⁴ + ...)*... =

= 1*1*1*1*...*1 + 1/2²*1*1*1*...*1 + 1/3²*1*1*1*...*1 + 1/5²*1*1*1*..*1 + .... + 1/2²*1/3²*1*..*1 + ...

We note that it equals all combinations of 1/p².

(8) Applying the Fundamental Theorem of Arithmetic (see here), this gives us:

1*1*1*1*...*1 + 1/2²*1*1*1*...*1 + 1/3²*1*1*1*...*1 + 1/5²*1*1*1*..*1 + .... + 1/2²*1/3²*1*..*1 + ...
=∑ (n=1,∞) 1/n²

(9) So putting it all together gives us:

∏ (p = all primes)1/(1 - 1/p²) = ∑ (n=1, ∞) 1/n²

(10) But, we know that:

∑ (n=1, ∞) 1/n² = π²/6 (see here)

(11) Now, taking the reciprocal of this gives us:

∏ (p = all primes)(1 - 1/p²) = 6/π².

QED

References

• The zeta function, odds that two numbers are coprime, MathReference.Com
  

Sum of the reciprocal of the primes

In today's blog, I will show that the sum of the reciprocal of the primes is divergent, that is, it does not have a limit.

This was first proved by Leonhard Euler. The details in today's blog are based on Euler's original proof.

Lemma 1: ∑ (p=all primes) ∑ (n=2,∞) (p^-n)/n is convergent, that is, it has a finite limit.

Proof:

(1) ∑ (p = all primes) ∑ (n=2,∞) (p^-n)/n is less than ∑ (p = all primes) ∑ (n=2,∞) (p^-n)

This is clear since n is an integer that starts at 2 and goes on until infinity. In each case, we have (p^-n) is greater than (p^-n)/n

(2) Using the infinite geometric series equation (see Lemma 1, here), we know that:

1 + 1/p + 1/p² + ... = 1/(1 - 1/p)

(3) Now (1/p²)*[1 + 1/p + 1/p² + 1/p³ + ... ] = [1/p² + 1/p³ + ..] so that:

∑ (n=2, ∞) (p^-n) = (1/p²)[1/(1 - 1/p)] = 1/[p²(1 - p^-1)]

(4) We can then use step #3 to give us:

∑ (p = all primes) ∑ (n=2, ∞) (p^-n) = ∑ (p = all primes) 1/[p²(1 - p^-1)] =

= ∑ (p = all primes) 1/[(p)(p)(1 - p^-1)] =

= ∑ (p = all primes) 1/[(p)(p - 1)]

(5) ∑ (p = all primes) 1/[(p)(p-1)] is less than ∑ (n=2, ∞) 1/[(n)(n-1)] since n includes nonprimes in the sum.

(6) ∑ (n=2, ∞) 1/[(n)(n-1)] = ∑ (n=2,∞) 1/(n² - n) is less than ∑ (n=2, ∞) 1/(n² - 2n + 1)

This is true since n² - 2n + 1 is always less than n² - n when n ≥ 2.

(7) Since n² - 2n + 1 = (n-1)², we have:

∑ (n=2,∞) 1/(n-1)² = 1 + 1/4 + 1/9 + ... = π²/6 (See Theorem, here)

(8) So, we have now proved that:

∑ (p = all primes) ∑ (n=2,∞) (p^-n)/n is less than π²/6 which is finite.

QED

Theorem: Sum of the Reciprocal of Primes is Divergent

∑ 1/p = 1/2 + 1/3 + 1/5 + 1/7 + ... is divergent, that is, does not have a limit.

Proof:

(1) We know that the ζ(1) = harmonic series, that is, ∑ 1/n diverges, that is, the limit of ∑ 1/n approaches infinity. [See Lemma 3, here]

NOTE: ζ(s) is the zeta function which is equal to ∑ 1/n^s. So ζ(1) = ∑ 1/n¹ = ∑ 1/n.

(2) From step #1, the limit of log(ζ(1)) approaches infinity. This follows directly from step #1 and the fact that as x gets bigger, log x likewise gets bigger. As x heads to infinity, log x also heads to infinity (see here for background on logarithms)

(3) Using Euler's Product Formula (see Theorem 4, here), we know that:

log(ζ(1)) = log(∑ 1/n) = log(∏ (1 - p^-1)^-1)

NOTE: ∏ (1 - p^-1)^-1) is a product of all primes p.

(4) By the property of logarithms (see here), we have:

log(∏ (1 - p^-1)^-1) = ∑ log([1 - p^-1]^-1) = ∑ -log(1 - p^-1)

(5) Now using some basic properties of the integral (see Theorem, here), we have:

log (1 + x) = ∑ (n=1,∞) [(-1)ⁿ⁺¹xⁿ]/n

Let x = -p^-1

Then we have:

log (1 + [-p^-1]) = ∑ [(-1)ⁿ⁺¹(-p^-1)ⁿ]/n = ∑ [(-1)ⁿ⁺¹(-1)ⁿ(p^-1)ⁿ]/n =

= ∑ [(-1)²ⁿ⁺¹(p^-n)]/n

Since 2n+1 is always odd, this gives us:

log(1 + [-p^-1]) = ∑ -(p^-n)/n

Since p^-n/n is always positive, we have ∑ -(p^-n)/n = (-1)∑ (p^-n)/n

Further,

-log(1 + [-p^-1]) = (-1)log(1 + [-p^-1]) = (-1)(-1)∑ (p^-n)/n = ∑ (p^-n)/n

(6) Combining step #5 with step #4 gives us:

∑ -log(1 - p^-1) = ∑ (p=all primes) ∑ (n=1,∞) (p^-n)/n

(7) ∑ (p=all primes) ∑ (n=1,∞) (p^-n)/n=∑(p=all primes) 1/p + ∑ ∑ (n=2,∞) (p^-n)/n

(8) Using Lemma 1 above, we know that ∑ ∑ (n=2,∞) (p^-n)/n is convergent, that is, it has a finite limit.

(9) But from step #2, we know that ∑ 1/p + ∑ ∑ (n=2, ∞) (p^-n)/n has a limit of infinity.

(10) Therefore, it follows, that ∑ 1/p must be divergent, that is, it must have a limit of infinity.

QED

References

• Jay R. Goldman, The Queen of Mathematics, A K Peters, 1998
• Proof that the Sum of the Reciprocals of Primes Diverge, Wikipedia
• Natural Logarithm, Wikipedia
  

∑ 1/n2 = π2/6

In today's blog, I show the proof that ∑ 1/n² = π²/6. For background on Basel's problem and Leonhard Euler's technique for arriving at the solution, see here. In today's blog, I show a proof that Euler's solution is correct.

Most of the content in today's blog is taken from the Wikipedia article on the Basel Problem. See references below for more information.

Lemma 1: cot²(x) is one-to-one on the interval (0,π/2)

Proof:

(1) Suppose there exists x,y such that cot²(x) = cot²(y) in the interval (0,π/2)

(2) By definition, cot(x) = (cos x)/(sin x) so that we have:

(cos² x)/(sin² x) = (cos² y)/(sin² y)

which means that:

(cos² x)/(cos² y) = (sin² x)/(sin² y)

(3) We also know (see Corollary 2, here), cos²(x) + sin²(x) = 1 which is the same as:

1 - sin²(x) = cos²(x)

(4) Now, we can make the same argument for y and then we can divide by y to get:

[1 - sin²(x)]/[1 - sin²(y)] = [cos²(x)]/[cos²(y)]

(5) Now, we can substitute in step #2 to give us:

[1 - sin²(x)]/[1 - sin²(y)] = (sin² x)/(sin² y)

(6) This amounts to:

(sin² y)(1 - sin² x) = (sin² x)(1 - sin² y)

(7) Adding (sin² x)(sin² y) to both sides gives us:

(sin² y)(1 - sin² x) + (sin² x)(sin² y) =

= (sin² y)(1 - sin² x + sin² x) = sin² y

(sin²x)(1 - sin² y) + (sin² x)(sin² y) =

= (sin²x)(1 - sin²y + sin² y) = sin² x

So that we have:

sin² y = sin² x

(8) Squaring both sides gives us:

±(sin y) = ±(sin x)

(9) Since we know that the sin function is nonnegative on the interval (0, π/2), we know that the square has to be nonnegative. This gives us:

sin y = sin x

(10) Finally, since sin x is one-to-one on the interval (0,π/2) (see here for properties of sin), we can see that x = y.

The basic idea here is that the sin function between 0 and π/2 does not repeat any of its values. This proves that if sin x = sin y, then x = y.

QED

Lemma 2: Sum of coefficients for a polynomial

if :

f(t) = a_mt^m + a_m-1t^m-1 + ... + a₁t + a₀ where a_m ≠ 0

then:

the sum of the roots of f(t) [counting multiplicities] is -a_m-1/a_m

Proof:

(1) Using the Fundamental Theorem of Algebra (see here), we know that we can divide up f(t) into a sequence of m roots such that:

f(t) = a_m(t - r₁)(t - r₂)*...*(t - r_m)

(2) Now when we multiply all these values together it is clear that the sum of coefficients for t^m-1 = the sum of all the roots multiplied by a_m.

We can see that if we carry out the multiplication of (t - r₁)(t - r₂)*...*(t - r_m), we get:

t^m + t^m-1(-r₁ + -r₂ + ... + -r_m) + ... + (-r₁)(-r₂)*...*(-r_m)

(3) We can see that the coefficient for t^m-1, a_m-1 = -a_m(r₁ + r₂ + ... + r_m)

(4) But then r₁ + r₂ + ... + r_m = -a_m-1/a_m

QED

Lemma 3: csc²(x) = 1 + cot²(x)

Proof:

(1) sin²x + cos²x = 1 (see Corollary 2, here)

(2) Divide both sides by sin²

Then we have:

1/(sin²x) = 1 + (cos²x)/(sin²x)

(3) Since by definition, cot²x = (cos² x)/(sin² x) and csc²x = 1/(sin²x), step #2 gives us:

csc² x = 1 + cot² x

QED

Lemma 4: cot²x is less than 1/x² which is less than csc² x on the interval (0,π/2)

Proof:

(1) If x is in (0, π/2), then (see Theorem, here):

sin x is less than x which is less than tan x.

(2) Taking the inverse of each (since x ≠ 0 → sin x ≠ 0 and tan x ≠ 0) gives us:

csc x is greater than 1/x which is greater than cot x.

(3) Now, squaring each value gives us:

cot² x is less than 1/x² which is less than csc² x.

QED

Lemma 5: as m approaches infinity, both (2m)(2m+2)/(2m+1)² and (2m)(2m-1)/(2m+1)² approaches 1

Proof:

(1) Putting all values into polynomial form, we have:

(2m)(2m+2) = 4m² + 4m

(2m)(2m - 1) = 4m² - 2m

(2m+1)² = 4m² + 4m + 1

(2) Now, we can calculate the limits using L'Hopital's rule (see here)

lim (m → ∞) [(4m² - 2m)/(4m² + 4m + 1)] = lim (m → ∞) [(8m - 2)/(8m + 4)] = lim (m → ∞) [8/8] = 1.

lim (m → ∞) [(4m² + 4m)/(4m² + 4m + 1)] = lim (m → ∞) [(8m + 4)/(8m + 4)] = 1.

QED

Theorem: ∑ 1/n² = π²/6

Proof:

(1) Using De Moivre's Formula (see Corollary, here for proof) we know that:

(cos x + i sin x)ⁿ = cos(nx) + i sin(nx)

(2) Dividing both sides by (sin x)ⁿ gives us:

[cos(nx) + i sin(nx)]/(sin x)ⁿ = (cos x + i sin x)ⁿ/(sin x)ⁿ = [(cos x + i sin x)/sin x]ⁿ

(3) Since cot x = (cos x)/(sin x) [by definition, cot(x) = 1/tan(x)], we have:

[(cos x + i sin x)/sin x] = cot x + i

(4) Putting this all together gives us:

[ cos(nx) + i sin(nx) ]/(sin x)ⁿ = (cot x + i)ⁿ

(5) Using the Binomial Theorem [See here], we know that:

(cot x + i)ⁿ = cotⁿ x + (n,1)(cot^n-1x)i + ... + (n,n-1)(cot x)i^n-1 + iⁿ

(6) Since i² = -1, we can now divide up all terms in step #5 so that we have:

cotⁿ x + (n,1)(cot^n-1x)i + ... + (n,n-1)(cot x)i^n-1 + iⁿ = [(cotⁿx) - (n,2)(cot^n-2 x) ± ... ] + i[(n,1)(cot^n-1x) - (n,3)(cot^n-3x) ± ... ]

(7) Since for any identity a + bi = c + di, we know that a=c and b=d, we can build the following equation from step #2 and step #6:

[isin(nx)]/(sin x)ⁿ = i[(n,1)(cot^n-1x) - (n,3)(cot^n-3x) ± ... ]

(8) Dividing both sides by i gives us:

sin(nx)/(sin x)ⁿ = [(n,1)(cot^n-1x) - (n,3)(cot^n-3x) ± ... ]

(9) Now, since step #8 is an identity, it is true for all values n,x.

Let n = 2m + 1 where m is a positive integer.
Let x = rπ/(2m + 1) where r can = 1, 2, ..., m

From this, we can see that:

nx = (2m + 1)*(rπ)/(2m+1) = rπ

(10) From a property of sin (see Lemma 1, here), we can see that:

sin(nx) =0

(11) Applying these values to step #8 gives us:

sin(nx)/(sin x)ⁿ = 0/(sin x)ⁿ = 0 =

= [(n,1)(cot^n-1x) - (n,3)(cot^n-3x) ± ... ] =

= [(2m+1,1)(cot^2mx) - (2m+1,3)(cot^2m-2x) ± ... + (-1)^m ]

(12) Using Lemma 1 above, we can conclude that the values for x in step #11 are distinct. By the above equation we know that each of these m distinct solutions for x are roots to the following equation:

(2m + 1, 1)t^m - (2m + 1,3)t^m-1 ± ... + (-1)^m = 0

(13) Using Lemma 2 to calculate the sum of roots using the coefficients, we have:

cot²(π/[2m+1]) + cot²(2π/[2m + 1]) + ... + cot²(mπ/[2m + 1]) = (2m+1,3)/(2m+1,1) = [(2m+1)!/(2m-2)!(3!)]/[(2m+1)!/(2m)!] = [(2m+1)(2m)(2m-1)/6]/(2m+1)] = (2m)(2m-1)/6.

(14) Using Lemma 3 above, we know that:

csc² x = cot²x + 1

which means that:

cot²x = csc² x - 1

(15) Applying (cot²x = csc²x - 1) to step #13 gives us:

cot²(π/[2m+1]) + cot²(2π/[2m + 1]) + ... + cot²(mπ/[2m + 1]) =

csc²(π/[2m+1]) + csc²(2π/[2m + 1]) + ... + csc²(mπ/[2m + 1]) - m =

= (2m)(2m-1)/6.

(16) If we add m to both sides we get:

csc²(π/[2m+1]) + csc²(2π/[2m + 1]) + ... + csc²(mπ/[2m + 1]) = (2m)(2m-1)/6 + m = (2m)(2m-1)/6 + 6m/6 = (4m² - 2m + 6m)/6 = (4m² + 4m)/6 = (2m)(2m+2)/6

(17) From Lemma 4 above, we know that for the interval (0,π/2) we have the following inequality:

cot²x is less than 1/x² which is less than csc² x

(18) But this implies that:

cot²(π/[2m+1]) + cot²(2π/[2m + 1]) + ... + cot²(mπ/[2m + 1]) is less than ([2m+1]/π)² + ([2m+1]/2π)² + ... + ([2m+1]/mπ)² which is less than csc²(π/[2m+1]) + csc²(2π/[2m + 1]) + ... + csc²(mπ/[2m + 1])

(19) Now since we know that

(a) cot²(π/[2m+1]) + cot²(2π/[2m + 1]) + ... + cot²(mπ/[2m + 1]) = (2m)(2m-1)/6

(b) csc²(π/[2m+1]) + csc²(2π/[2m + 1]) + ... + csc²(mπ/[2m + 1]) = (2m)(2m+2)/6

We have:

(2m)(2m-1)/6 is less than ([2m+1]/π)² + ([2m+1]/2π)² + ... + ([2m+1]/mπ)² which is less than (2m)(2m+2)/6

(20) If we multiply on all sides by (π/(2m+1))², then we have:

π²/(2m+1)²(2m)(2m-1)/6 = (π²)(2m)(2m-1)/(6)(2m+1)²

(π²)/(2m+1)²([2m+1]/π)² + ([2m+1]/2π)² + ... + ([2m+1]/mπ)² = 1/1 + 1/4 + .. + 1/m²

(π²)/(2m+1)²(2m)(2m+2)/6 = π²(2m)(2m-1)/(6)(2m+1)²

(21) By Lemma 5 above, as m approaches infinity, both the left and right hand expressions approach (π²)/6 so by the Squeeze Theorem (see Lemma 3, here for proof), we have:

∑ (k=1,∞) (1/k²) = lim (m → ∞) (1/1² + 1/2² + ... + 1/m²) = π²/6

QED

References

• Basel Problem, Wikipedia