Zeta Infinite Summation $\sum_{n=1}^{\infty}(-1)^{n-1}\,\zeta(s+n)$

Question

Let $Re\{s\}\gt0$ : $$ \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} = \sum_{n=1}^{\infty} \left( \frac{n}{n+1}\right) n^{-(s+1)} = \sum_{n=1}^{\infty} \left( 1 - \frac{1}{n+1}\right) n^{-(s+1)} = \zeta(s+1) - \sum_{n=1}^{\infty} \frac{n^{-(s+1)}}{n+1} = \\[6mm] \zeta(s+1) - \zeta(s+2) + \sum_{n=1}^{\infty} \frac{n^{-(s+2)}}{n+1} = \zeta(s+1) - \zeta(s+2) + \zeta(s+3) - \sum_{n=1}^{\infty} \frac{n^{-(s+3)}}{n+1} = \text{...} \Rightarrow \\[6mm] \sum_{n=1}^{N} (-1)^{n-1} \zeta(s+n) = \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} - (-1)^{N}\sum_{n=1}^{\infty} \frac{n^{-(s+N)}}{n+1} \Rightarrow \\[6mm] \boxed{ \quad \sum_{n=1}^{\infty} (-1)^{n-1} \zeta(s+n) = \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} - \lim_{N\rightarrow\infty}\left[(-1)^{N}\sum_{n=1}^{\infty} \frac{n^{-(s+N)}}{n+1}\right] \quad } \\[6mm] $$

Does the limit exist? and What does it equal? $$ L = \lim_{N\rightarrow\infty}\left[(-1)^{N}\sum_{n=1}^{\infty} \frac{n^{-(s+N)}}{n+1}\right] \qquad\qquad\colon\space Re\{s\} \gt 0 \tag{1}$$ $$\sum_{n=1}^{\infty} (-1)^{n-1} \zeta(s+n) = \sum_{n=1}^{\infty} \frac{n^{-s}}{n+1} - L \qquad\colon\space Re\{s\} \gt 0 \tag{2}$$

Without the outer sign, the limit is: $$ \small \sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n^2}\lt\sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n+1}\lt\sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n} \Rightarrow \zeta(s+N+2)-1 \lt \sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n+1} \lt \zeta(s+N+1)-1 \\ \small \text{Let}\space\left\{N\rightarrow\infty\right\}\space\text{and use the limit}\space\left\{\lim_{z\rightarrow\infty}\zeta(z)=1\right\} \Rightarrow \lim_{N\rightarrow\infty}\sum_{n=2}^{\infty}\frac{n^{-(s+N)}}{n+1}=0 \Rightarrow \color{red}{\lim_{N\rightarrow\infty}\sum_{n=1}^{\infty}\frac{n^{-(s+N)}}{n+1}=\frac{1}{2}} \\ $$ NB: appreciating your explanations on a similar previous question. Many Thanks.

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conclusion:

As of the correct answer(s): $$ \sum_{n=1}^{\infty}\frac{n^{-s}}{n+1} - \sum_{n=1}^{\infty}(-1)^{n-1}\,\zeta(s+n) = \sum_{n=1}^{\infty}\frac{n^{-s}}{n+1} - \sum_{n=1}^{\infty}\left[\color{red}{\zeta(s+2n-1)-\zeta(s+2n)}\right] \\[6mm] \quad = \lim_{N\rightarrow\infty}\sum_{n=1}^{\infty}\frac{n^{-(s+N)}}{n+1} = \color{red}{\frac{1}{2}} \quad\colon\space Re\{s\}\ge0 \quad\{\small\text{holds for s=0 too}\normalsize\} \\[6mm] $$

Answer 1

Using the geometric series, for $Re(s) > 0$ $$\sum_{n=1}^\infty \frac{n^{-s}}{n+1} = \frac12+\sum_{n=2}^\infty n^{-s-1}\sum_{k=0}^\infty n^{-k}(-1)^k = \frac12+\sum_{k=0}^\infty\sum_{n=2}^\infty n^{-s-1} n^{-k}(-1)^k$$ $$ = \frac12+\sum_{k=0}^\infty (-1)^k (\zeta(s+k+1)-1)= \frac12+\sum_{k=1}^\infty (\zeta(s+2k-1)-\zeta(s+2k))$$ where the change of order of summation is allowed because $\sum_{n=2}^\infty |n^{-s-1}|\sum_{k=0}^\infty n^{-k}= \sum_{n=2}^\infty \frac{n^{-Re(s)-1}}{1-\frac1n}$ so it converges absolutely

Answer 2

There are two cases:

$$(-1)^N\sum_{n=1}^\infty\frac{n^{-s+N}}{n+1}=\begin{cases}\ \ \ \sum_{n=1}^\infty\frac{n^{-s+N}}{n+1}\\-\sum_{n=1}^\infty\frac{n^{-s+N}}{n+1}\end{cases}$$

For the limit to exist, the positive limit must equal the negativet limit, and thus we get that $L$ must be $0$ if it exists. It is trivial to show that the limit cannot be zero.

Answer 3

Let $S(N) =\sum_{n=1}^{N} \frac{n^{-s}}{n+1} =\sum_{n=1}^{N} \frac{1}{n^s(n+1)} $.

If $Re(s)> 0$, $\lim_{N \to \infty} S(N)$ exists by comparison with $\sum_{n=1}^{N} \frac{1}{n^{1+Re(s)}} $.

Then, since $\frac1{1+x} =\sum_{k=0}^{2m} (-1)^k x^k -\frac{x^{2m+1}}{1+x} $

$\begin{array}\\ S(N) &=\sum_{n=1}^{N} \frac{1}{n^s(n+1)}\\ &=\sum_{n=1}^{N} \frac{1}{n^{s+1}(1+1/n)}\\ &=\sum_{n=1}^{N} \frac{1}{n^{s+1}}(\sum_{k=0}^{2m} (-1)^k n^{-k} -\frac{n^{-2m-1}}{1+1/n})\\ &=\sum_{n=1}^{N} \frac{1}{n^{s+1}}\sum_{k=0}^{2m} (-1)^k n^{-k} -\sum_{n=1}^{N} \frac{1}{n^{s+1}}\frac{n^{-2m-1}}{1+1/n}\\ &=\sum_{n=1}^{N} \sum_{k=0}^{2m} (-1)^k \frac1{n^{k+s+1}} -\sum_{n=1}^{N} \frac{1}{(1+1/n)n^{s+2m+2}}\\ &= \sum_{k=0}^{2m}\sum_{n=1}^{N} (-1)^k \frac1{n^{k+s+1}} -\sum_{n=1}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\ &= \sum_{k=0}^{2m}(-1)^k\sum_{n=1}^{N} \frac1{n^{k+s+1}} -\sum_{n=1}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\ &= S_1(N)-S_2(N)\\ \end{array} $

As $N \to \infty$, $S_1(N) = \sum_{k=0}^{2m}(-1)^k\sum_{n=1}^{N} \frac1{n^{k+s+1}} \to \sum_{k=0}^{2m}(-1)^k \zeta(k+s+1) $ and

$\begin{array}\\ S_2(N) &=\sum_{n=1}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\ &\gt \frac12\\ \text{and}\\ S_2(N) &=\frac12+\sum_{n=2}^{N} \frac{1}{(n+1)n^{s+2m+1}}\\ &\le\frac12+\sum_{n=2}^{N} \frac{1}{(n+1)2^{2m}n^{s+1}}\\ &=\frac12+\frac1{4^m}\sum_{n=2}^{N} \frac{1}{(n+1)n^{s+1}}\\ &\lt\frac12+\frac1{4^m}\sum_{n=2}^{N} \frac{1}{(n+1)n}\\ &=\frac12+\frac1{4^m}\sum_{n=2}^{N} (\frac1{n}-\frac{1}{n+1})\\ &<\frac12+\frac1{2\cdot 4^m}\\ \end{array} $

Therefore $\lim_{m \to \infty} S_2(N) = \frac12 $, so that $\lim_{m \to \infty, N \to \infty} S(N) =-\frac12+\sum_{k=0}^{2m}(-1)^k \zeta(k+s+1) $.

Note that, if we write $\sum_{k=0}^{2m}(-1)^k \zeta(k+s+1) =\sum_{k=0}^{m-1}(\zeta(2k+s+1)-\zeta(2k+s+2)) +\zeta(2m+s+1) $, the sum converges properly. If we do this grouping earlier on (even and odd terms together with opposite signs), this all becomes rigorous.