The Riemann Zeta Function

The Riemann zeta function is defined by the p-series

ζ(p) = ∑_{n=1}^{∞} \frac{1}{n^{p}} = 1 + \frac{1}{2^{p}} + \frac{1}{3^{p}} + \frac{1}{4^{p}} + …, valid for p > 1,  (1)

which converges for p > 1 by the Integral Test (and diverges for p ≤ 1). One interesting special value [though hard to prove] is

ζ(2) = ∑_{n=1}^{∞} \frac{1}{n^{2}} = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \frac{1}{5^{2}} + … = \frac{π^{2}}{6} .  (2)

Convergence Suppose we try to use equation (2) to compute π^{2}/6 with an error of less than 10^{−6}. Question: How many terms do we need? Explicitly, how large must k be to bring ∑_{n=1}^{k} 1/n^{2} within 10^{−6} of π^{2}/6?

We note first that 1/n^{2} < 10^{−6} whenever n > 1000. This suggests taking k = 1000. But consider the next thousand terms, each of which is at least 1/2000^{2}; their sum is therefore greater than 1000 × 1/2000^{2} = 0.000250, which is far larger than 10^{−6}. [The error is actually 0.0009995, according to maple. In fact, maple knows about the series equation (1), and the remainder after k terms is a built-in function.]

The correct answer is (exactly) 1,000,000, with error 9.999995 × 10^{−7}. Thus equation (1) is highly impractical for computation.

An alternating series We introduce the variant

f(p) = ∑_{n=1}^{∞}(−1)^{n−1} \frac{1}{n^{p}} = 1 − \frac{1}{2^{p}} + \frac{1}{3^{p}} − \frac{1}{4^{p}} + …, valid for p > 0,  (3)

which converges for p > 0 by the Alternating Series Test.

Rearrangement The function f(p) is easily expressed in terms of ζ(p), when p > 1. Consider the odd and even terms separately, by defining

g(p) = 1 + \frac{1}{3^{p}} + \frac{1}{5^{p}} + \frac{1}{7^{p}} + …, valid for p > 1,

and

h(p) = \frac{1}{2^{p}} + \frac{1}{4^{p}} + \frac{1}{6^{p}} + …, valid for p > 1.

Clearly, g(p)+h(p) = ζ(p). This rearrangement is valid for p > 1. Also, multiplication of each term of h(p) by 2^{p} gives 2^{p}h(p) = ζ(p).

We have enough information to deduce that

h(p) = \frac{1}{2^{p}} ζ(p),  g(p) = \frac{2^{p} − 1}{2^{p}}ζ(p) .  (4)

Then f(p) = g(p) − h(p) gives

f(p) = (\frac{2^{p} − 1}{2^{p}} − \frac{1}{2^{p}} ) ζ(p) = \frac{2^{p} − 2}{2^{p}}ζ(p) .  (5)

More examples For p = 2, equation (4) gives two more well-known series,

g(2) = 1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + … = \frac{3}{4}ζ(2) = \frac{π^{2}}{8}  (6)

and

f(2) = 1 − \frac{1}{2^{2}} + \frac{1}{3^{2}} − \frac{1}{4^{2}} + … = \frac{π^{2}}{12} .  (7)

By Theorem 8, we only need 1000 terms of the series equation (7) to guarantee that the error is less than 10^{−6}. [The actual error is only 4.997 × 10^{−7}.] This is a big improvement over equation (2), but still hardly practical.

Extension Nevertheless, equation (3) is useful for more than just computation. We can turn equation (5) around to the form

ζ(p) = \frac{2^{p}}{2^{p} − 2}f(p) for p > 1.  (8)

However, the right side of this equation is defined for all p > 0 (provided we exclude the case p = 1, so as not to divide by 0). This suggests that it is reasonable to extend the definition of ζ(p) to all p > 0 (p ≠ 1) by means of this equation. [This is in fact true, but requires more justification than this.]