Sums of Powers of Positive Integers

[lpetrich]

Let's consider sums of powers of positive integers.
\( \sum_{k=1}^n 1 = n \\ \sum_{k=1}^n k = \frac12 n (n+1) \\ \sum_{k=1}^n k^2 = \frac16 n (n+1) (2n+1) \\ \sum_{k=1}^n k^3 = \frac14 n^2 (n+1)^2 \\ \sum_{k=1}^n k^4 = \frac{1}{30} n (n+1) (2n+1) (3n^2 + 3n - 1) \)

Not very much of a pattern. Let's look at something similar.

\( \sum_{k=1}^n k (k+1) \cdots (k+m) = \frac{1}{m+2} n (n+1) \cdots (n+m+1) \)

Easy to prove. But Abramowitz and Stegun have some general formulas for positive-integer powers, though they use something called Bernoulli polynomials, B_n(x). These are defined by

\( \frac{t e^{x t}}{e^t - 1} = \sum_{n=0}^{\infty} B_n(x) \frac{t^n} {n!} \)

Bernoulli numbers B_n are B_n(0). Some related polynomials are the Euler polynomials:

\( \frac{2e^{x t}}{e^t + 1} = \sum_{n=0}^{\infty} E_n(x) \frac{t^n} {n!} \)

However, the corresponding Euler numbers are E_n = 2ⁿ E_n(1/2).

The sum-of-powers formula:

\( \sum_{k=0}^n k^m = \frac{ B_{m+1}(n+1) - B_{m+1} }{m+1} \)

 

[lpetrich]

Bernoulli and Euler numbers fill out the series expressions for the trigonometric functions. The best-known series are

\( \sin x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} \\ \cos x = \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!} \)

For the remaining four functions, the series are
\( \tan x = \sum_{k=0}^\infty (-1)^{k+1} 2^{2k} (2^{2k}-1) B_{2k} \frac{x^{2k-1}}{(2k)!} \\ \cot x = \sum_{k=0}^\infty (-1)^{k} 2^{2k} B_{2k} \frac{x^{2k-1}}{(2k)!} \\ \sec x = \sum_{k=0}^\infty (-1)^k E_{2k} \frac{x^{2k}}{(2k)!} \\ \csc x = \sum_{k=0}^\infty (-1)^{k+1} 2 (2^{2k-1}-1) B_{2k} \frac{x^{2k-1}}{(2k)!} \\ \)

For products and powers of trigonometric functions, one can use trigonometric identities and differentiation, like: cos(x)² = (1/2)*(1 + cos(2x)) and sec(x)² = (d/dx) tan(x).

 

[lpetrich]

Negative powers of positive integers have some interesting features. Let's consider reciprocals first.

\( \sum_{k=1}^n \frac{1}{k} \to \log n + \gamma \)

as n -> oo, where the last term is the Euler-Mascheroni constant, 0.577216...

It can be generalized to

\( \sum_{k=1}^n \frac{(\log k)^m}{k} \to \frac{(\log n)^{m+1}}{m+1} + \gamma_m \)

For negative powers less than -1, the series will converge, and we get something called the "Riemann zeta function" in the power:

\( \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p} \frac{1}{1 - p^{-s}} = \frac{1}{\Gamma(s)} \int_{x=0}^\infty \frac{x^{s-1}}{e^x - 1} dx \)

where the p's are all the prime numbers. There are some interrelationships, like

\( \sum_{n=0}^\infty \frac{1}{(2n+1)^s} = (1 - 2^{-s}) \zeta(s) \\ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} = (1 - 2^{1-s}) \zeta(s) = \frac{1}{\Gamma(s)} \int_{x=0}^\infty \frac{x^{s-1}}{e^x + 1} dx \)

 

[lpetrich]

The Riemann zeta function has the series expansion

\( \zeta(s) = \frac{1}{s-1} + \sum_{n=0}^\infty (-1)^n \gamma_n \frac{(s-1)^n}{n!} \)

It diverges for s = 1, as one might expect, but for s < 1, the series will still converge, despite its defining sum failing to converge. This is from something called "analytic continuation", something which is roughly going around poles or divergent spots like s = 1 here, going around them in the complex plane.

One finds a remarkable sort of reflection theorem:

\( \zeta(s) = 2^s \pi^{s-1} \sin ((1/2) \pi s) \Gamma(1-s) \zeta(1-s) \)

It gives values of the function for s < 1, values where the function's series diverges. Some special values, for positive integers n:

\( \zeta(0) = \frac12 ,\ \zeta(1) = \infty ,\ \zeta(-2n) = 0 ,\ \zeta(1-2n) = - \frac{B_{2n}}{2n} ,\\ \zeta(2n) = \frac{(2\pi)^{2n}}{2 (2n)!} (-1)^{n+1} B_{2n} \)

Another special value:
\( \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \log 2 \)

 

[lpetrich]

It is evident that the Riemann zeta function has zeros -2, -4, -6, ..., its "trivial" ones. There is a theorem about its remaining zeros, its "nontrivial" ones. It is that these zeros all have real part 1/2 -- the  Riemann hypothesis This theorem has never been proved, despite a lot of effort and despite failure to find counterexamples. It has implications in, among other things, the distribution of prime numbers.

 

[lpetrich]

Thus, one gets

\( \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} \\ \sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90} \)

 

[lpetrich]

 Riemann zeta function has some more stuff about this function. Like a symmetric version:

\( \xi(s) = \frac12 \pi^{-s/2} s (s-1) \Gamma(s/2) \zeta(s) \)

It satisfies \( \xi(s) = \xi(1 - s) \)

There are ways of constructing series expansions that converge over smaller values of the real part of the arg. The defining series converges only for Re(s) > 1, and this series converges for Re(s) > 0:

\( \zeta(s) = \frac{1}{1 - 2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} \)

and these "Dirichlet series", converging for Re(s) > 0 and Re(s) > -1:

\( \zeta(s) = \frac{1}{s-1} \sum_{n=1}^\infty \left( \frac{n}{(n+1)^s} - \frac{n - s}{n^s} \right) \)

\( \zeta(s) = \frac{1}{s-1} \sum_{n=1}^\infty \frac{n(n+1)}{2} \left( \frac{2n+3+s}{(n+1)^{s+2}} - \frac{2n - 1 - s}{n^{s+2}} \right) \)

 

[lpetrich]

But there is an interesting series for the reciprocal of the Riemann zeta function:

\( \frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s} \)

where μ is the Möbius (Moebius) mu function. Defined for positive-integer n, it has values

• n = 1: 1
• n has maximum power 1 of all its primes (it's square-free): (-1)^(number of primes)
• n's factors have at least one prime power greater than 1: 0

Thus, mu(1) = 1, mu(2) = -1, mu(3) = -1, mu(4) = 0, mu(5) = -1, mu(6) = 1, ...

-

This function appears in an interesting context. An algebraic field is a combination of additionlike and multiplicationlike operations on some set where both are commutative (abelian) groups, the first over the entire set, and the second over all the set but the additive identity (0).

All the finite fields are known, and they were discovered by Evariste Galois. They all have numbers of elements that are powers of primes: GF(p^n) for prime p. The fields GF(p) are integers 0, 1, ..., p-1 with addition and multiplication both modulo p. The fields GF(p^n) for n > 1 are more complicated.

Their elements are polynomials in some variable with coefficients in GF(p). Thus, GF(4) has elements 0, 1, x, 1+x for variable x. Addition and multiplication are defined in the usual ways for polynomials, though polynomial multiplication involves taking the remainder from division by some degree-n irreducible monic polynomial. "Irreducible" meaning that it cannot be factored, and monic meaning that the highest term has coefficient 1. Let us see what degree-2 irreducible monic polynomials are available for GF(4).

x^2 = x*x (no), x^2 + 1 = (x+1)^2 (no), x^2 + x = x*(x+1) (no), and x^2 + x + 1 (yes)

For the second one, remember that the coefficients are in GF(2), and 1 + 1 = 0 in GF(2).

For larger fields, one has several such polynomials available, and how many there are is given by this formula:

\( \frac1n \sum_{d | n} \mu(d) p^{n/d} \)

for all d evenly dividing n.

 

[lpetrich]

For GF(8), one has
x^3 (no), x^3 + 1 = (x^2+x+1)*(x+1) (no), x^3 + x = x*(x+1)^2 (no), x^3 + x + 1 (yes), x^3 + x^2 = x^2*(x+1) (no), x^3 + x^2 + 1 (yes), x^3 + x^2 + x = x*(x^2+x+1) (no), x^3+x^2+x+1 = (x+1)^3 (no)

Two irreducible ones.

For GF(9), one has

x^2 (no), x^2 + 1 (yes), x^2 + 2 = (x+1)*(x+2) (no), x^2 + x = x*(x+1) (no), x^2 + x + 1 = (x+2)^2 (no), x^2 + x + 2 (yes), x^2 + 2x = x*(x+2) (no), x^2 + 2x + 1 = (x+1)^2 (no), x^2 + 2x + 2 (yes)

Three irreducible ones.

 

[Swammerdami]

Thus, one gets

\( \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\)

This is the identity that propelled a young Leonhard Euler into stardom! His derivation is very simple but lets me practice my rusty LaTeX!

Start with two equations for sin(x); the first (given by Ipetrich above) was well-known by Euler's time; the second follows from the fact that a polynomial can be expressed via its zeros. (The zeros of sin(x) are 0, pi, 2 pi, etc.)

\(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... = K x (1 - \frac{x}{\pi})(1 + \frac{x}{\pi}) (1 - \frac{x}{2\pi})(1 + \frac{x}{2\pi}) (1 - \frac{x}{3\pi})(1 + \frac{x}{3\pi}) ... \)

The x^1 term on the left has coefficient 1, so K must be 1 in r.h.s. The r.h.s can be further simplified since (1+a)(1-a) = 1 - a^2.

\(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... = x (1 - (\frac{x}{\pi})^2) (1 - (\frac{x}{2\pi})^2) (1 - (\frac{x}{3\pi})^2) ... \)

Divide both sides by x.

\(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + ... = (1 - (\frac{x}{\pi})^2) (1 - (\frac{x}{2\pi})^2) (1 - (\frac{x}{3\pi})^2) ... \)

We're equating two polynomials so we can drop everything but the (- x^2) terms.

\( ... \frac{x^2}{6} + ... = ... + ((\frac{x}{\pi})^2 + (\frac{x}{2\pi})^2 + (\frac{x}{3\pi})^2) + ...) + ... \)

Multiply everywhere by \((\frac{\pi}{x})^2\)

\( \frac{\pi^2}{6} = (1)^2 + (\frac{1}{2})^2 + (\frac{1}{3})^2 + ... \)

Q.E.D.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I saw a wonderful alternative proof of this Euler's Basel Identity on Youtube. It starts with the following elegant fact and its proof:

Place N lamps of uniform candle-power equally spaced along the circumference of a circular lake of diameter N. Place yourself along the circumference midway between two of the lamps. When N=1 (and you are directly opposite the single lamp) you receive 1^(-2) light, due to an inverse-square law. When N=2 you receive sqrt(2)^(-2) light from each lamp for the same total light.
The total light you receive is always exactly 1, regardless of N.​

When you work out the case \(N \rightarrow \infty\) (and note that an infinitesimal portion of a circle can be treated as a line!) you derive

\(\frac{\pi^2}{8} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ...\)

This identity can be manipulated into the usual Basel Identity.