Question

$\begin{gathered}\footnotesize \sf If \: \sum\limits_{n=1}^\infty \: \dfrac{1}{n^z} = 0 \: where, z\in{\mathbb{C}} \: and \: z\not=-2a \: \forall \: a\in N \\ \footnotesize \sf then \: find \: the \: value \: of \: {\mathfrak{Re}}(z)\end{gathered}$​

Answer 1

We can use the Euler-Maclaurin summation formula to approximate the value of the series for $\rm{\mathfrak{Re}}(z)>1$. The formula states that for a function f(x) and integers a, b such that a<b and h=b-a, we have:

$\rm \sum_{n=a}^{b} f(n) = \int_a^b f(x) \, dx + \frac{1}{2}(f(a)+f(b)) + \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} \left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right)$

where $\rm B_{2k}$ are the Bernoulli numbers and $\rm f^{(n)}(x)$ denotes the n-th derivative of f(x).

In our case, we have $\rm f(n) = \frac{1}{n^z}$ and a=1, b=N, so we can write:

$\rm \sum_{n=1}^{N} \frac{1}{n^z} = \int_1^N \frac{1}{x^z} \, dx + \frac{1}{2}(1+N^{-z}) + \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} \left(\frac{(-1)^{2k-1}(z)}{(2k-1)!}N^{-z-2k+1} - \frac{(-1)^{2k-1}(z)}{(2k-1)!}1^{-z-2k+1}\right)$

where we have used the fact that

$\rm f^{(n)}(x) = (-1)^{n-1} z(z+1)\cdots(z+n-1) x^{-z-n}.$

Taking the limit as $\rm N\rightarrow\infty$, the integral term goes to zero if $\rm{\mathfrak{Re}}(z)>1$, and the series converges if $\rm{\mathfrak{Re}}(z)>-1.$ Therefore, we can write:

$\rm 0 = \frac{1}{2} + \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} \left(\frac{(-1)^{2k-1}(z)}{(2k-1)!} 0 - \frac{(-1)^{2k-1}(z)}{(2k-1)!}1^{-z-2k+1}\right)$

which simplifies to:

$\rm0 = \frac{1}{2} + \sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!} \frac{(-1)^{2k-1}(z)}{(2k-1)!}$

This is an infinite series in z, but we can use the fact that the Bernoulli numbers satisfy the generating function:

$\rm \frac{t}{e^t-1} = \sum_{k=0}^{\infty} \frac{B_k}{k!} t^k$

to write:

$\rm 0 = \frac{1}{2} + \frac{t}{e^t-1} \frac{z}{t}$

where t=-2π i. This equation can be solved for z as:

$\rm z = \frac{2\pi i}{e^{2\pi i}-1} = \frac{2\pi i}{1} = 2\pi i$

Therefore, ${\mathfrak{Re}}(z) = 0$.

Note that this solution agrees with the fact that the series converges for ${\mathfrak{Re}}(z)>-1$, since the series diverges for ${\mathfrak{Re}}(z)\leq -1$.