Calculating values of the Riemann Zeta Function

The Riemann Zeta Function is most commonly defined as $$\zeta(s)=\sum_{n=0}^\infty \frac{1}{n^s}$$ There is some sort of million dollar prize that involves proving the real part of complex number s must be $\frac{1}{2}$ for all nontrivial zeros. Of course this intregued me, because well, it's a million dollars. Odds are I won't solve it, but still. Anyway, I started looking at it and realized that you'd be raising a number to a complex power. This made no sense to me, so I went online and found Euler's formula that explains how that would work $$e^{i\pi}=-1$$ It turns out that the smallest nontrivial zeros is at about $\frac{1}{2}+14.1345i$, so I plugged it in to the zeta function. I used Desmos.com, and used separate summations for the real and imaginary parts. I expected to get zero. I did not get zero. In fact, the bigger I had the summation get, say, instead of summing to 1000000, I'd sum to 1000000000, the further off I would get from zero.

So tell me, how exactly are values for the Riemann Zeta Function computed?

• That formula for the zeta function works only when $\text{Re}(s) > 1$. For other values of $s \neq 1$ on the complex plane, we need to use it's analytic continuation formula $\zeta(s) = 2^s \pi^{s-1} \sin\left(\dfrac{\pi s}{2}\right) \Gamma(1 - s) \zeta(1 - s)$ – Henrique Augusto Souza Feb 20 '17 at 14:00
• Actually, this formula: $\xi(s) = \pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)$. The $\xi(s)$ satisfies $\xi(s) = \xi(1 - s)$ and is defined everywhere except $s = 1$. – Henrique Augusto Souza Feb 20 '17 at 14:05
• Try this: en.wikipedia.org/wiki/… – Simply Beautiful Art Feb 20 '17 at 14:07
• Oh.... hold up sir. Please don't mark my answer as accepted, as you will likely receive more answers soon. – Simply Beautiful Art Feb 20 '17 at 14:29
• @HenriqueAugustoSouza That reflection formula doesn't reach $\zeta(s)$ for $0\le\Re(s)\le1$ – Simply Beautiful Art Feb 20 '17 at 14:37

One may note that when $\Re(s)\le1$,

$$\sum_{k=1}^\infty\frac1{k^s}\approx\int_1^\infty\frac1{x^s}\ dx\to\infty$$

Thus, we'll need a different representation of the zeta function. If we let $\eta(s)$ be the alternating form of the zeta function,

$$\eta(s)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^s}$$

Then,

$$\zeta(s)-\eta(s)=\sum_{k=1}^\infty\frac{1+(-1)^k}{k^s}=\sum_{k=1}^\infty\frac2{(2k)^s}=2^{1-s}\zeta(s)$$

Thus, it follows that

$$\zeta(s)=\frac1{1-2^{1-s}}\eta(s)$$

By taking the Euler sum of the $\eta$ function, we get

$$\eta(s)=\sum_{n=0}^\infty\frac1{2^{1+n}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}$$

and thus, we reach a globally convergent form of the Riemann zeta function:

$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\frac1{2^{1+n}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}$$

and when testing for zeroes, the $\frac1{1-2^{1-s}}$ part is negligible.

1 Answer cannot be correct. The author claims that his result is globally convergent. But it is not globally convergent to the zeta function. It is known that zeta(0)=-1/2 However if we set s=0 in his/her result the inner (second) sum reduces to the sum from k=0 to n of n!(-1)^k/((n-k)!(k)!) which is zero (by the binomial theorem). You can reply at adwunsch@gmail.com

• I withdraw my comment . The k=0, n=0 term gives the right answer. – ADavid Wunsch May 5 at 22:00