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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.

conclusion:

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

Community
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Hazem Orabi
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    $$\sum_{n=1}^{+\infty}(-1)^{n-1}\zeta(s+n)$$ is not a convergent series in the usual sense, since $\zeta(s+n)\approx 1$ for large values of $n$ (or $s$, or both) and $\sum_{n\geq 1}(-1)^{n-1}$ is not convergent in the usual sense. Are you considering such series "à la Cesàro"? – Jack D'Aurizio Dec 07 '16 at 22:48
  • @Jack D'Aurizio: Yes indeed! We are on the same page. Thank you. Any possibility to apply average Cesàro in this case? – Hazem Orabi Dec 07 '16 at 22:56
  • In such a case, by the integral representation of the $\zeta$ function, $$\sum_{n\geq 1}(-1)^{n-1}\zeta(s+n) = \int_{0}^{+\infty}\left(\sum_{m\geq 0}\frac{x^m}{(s+m)!}\right)\frac{x^s ,dx}{e^x-1}.$$ – Jack D'Aurizio Dec 07 '16 at 23:06
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    Grouping the terms by two $$\sum_{n \ge 1} \zeta(s+2n-1)-\zeta(s+2n)$$ converges absolutely. – reuns Dec 08 '16 at 11:03

3 Answers3

2

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

reuns
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  • Re-writing alternating terms as subtracting of odd and even indexes terms is a key to change divergent series to absolute convergent. Thanks. I am accepting your answer. – Hazem Orabi Dec 09 '16 at 00:16
0

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.

Simply Beautiful Art
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  • Would you mind to demonstrate how L cannot be zero? – Hazem Orabi Dec 07 '16 at 23:05
  • @HazemOrabi Add the first few terms. Where is it going? – Simply Beautiful Art Dec 07 '16 at 23:11
  • Because of the outer sign, we are facing a bit difficulties with the first term! not to mention few terms after. – Hazem Orabi Dec 07 '16 at 23:16
  • @HazemOrabi Which is why my answer splits the cases so we may ignore the outer sign. Now, what are the first few terms of the sum when the outer sign is positive? – Simply Beautiful Art Dec 07 '16 at 23:19
  • Sorry to bother you, But if we consider Jack’s comments regarding $\zeta(s+n)\approx 1$ and the last value of the limit without the outer sign, then the limit may exist (not in the usual sense), and may equal $\log(2)/2$ or $1/4$. Nevertheless, I have to admit, it may not exist. – Hazem Orabi Dec 07 '16 at 23:29
  • @HazemOrabi It clearly does not exist in the usual sense, since it fails the term test. Also, pretty much all sums exist outside of the usual sense, though I'm not sure why you'd bring that in. – Simply Beautiful Art Dec 07 '16 at 23:31
  • @HazemOrabi Also, I don't think you Cesàro summed that correctly, since it should be dependent on $s$. – Simply Beautiful Art Dec 07 '16 at 23:33
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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.

marty cohen
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  • Thank you for the correct answer. I allowed myself to choose a second correct answer. Please accept my apology. Meanwhile, may I ask your help in the (similar previous question) mentioned above. Appreciating. Hazem. – Hazem Orabi Dec 09 '16 at 00:16