Question in the title. Does $\sum_{n = 2}^{\infty} [\zeta(n) - 1]$ converge? If not, how about $\sum_{n = 1}^{\infty} [\zeta(2n) - 1]$?
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2$\begingroup$ Interestingly, wolframalpha says the solution is $1$: wolframalpha.com/input/… Also Wikipedia: en.wikipedia.org/wiki/… $\endgroup$ – Simply Beautiful Art Sep 16 '16 at 22:04
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$\begingroup$ Related: math.stackexchange.com/questions/1433797 $\endgroup$ – Watson Jun 17 '17 at 15:01
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About the first series:
$$ S=\sum_{n\geq 2}\left(\zeta(n)-1\right)=\int_{0}^{+\infty}\sum_{m\geq 1}\frac{x^m}{m!}\left(\frac{1}{e^x-1}-\frac{1}{e^x}\right)\,dx $$ leads to: $$ S = \int_{0}^{+\infty}(e^x-1)\left(\frac{1}{e^x-1}-\frac{1}{e^x}\right)\,dx = \int_{0}^{+\infty}e^{-x}\,dx = \color{red}{1}.$$
About the second series:
$$S= \sum_{n\geq 1}\left(\zeta(2n)-1\right) = \int_{0}^{+\infty}\sum_{m\geq 0}\frac{x^{2m+1}}{(2m+1)!}\left(\frac{1}{e^x-1}-\frac{1}{e^x}\right)\,dx$$ hence $$ S = \int_{0}^{+\infty}\frac{\sinh(x)}{e^x(e^x-1)}\,dx = \frac{1}{2}\int_{1}^{+\infty}\frac{u-\frac{1}{u}}{u^2(u-1)}\,du=\frac{1}{2}\int_{1}^{+\infty}\frac{u+1}{u^3}\,du=\color{red}{\frac{3}{4}}. $$ The same technique (i.e. to exploit the integral representation for the $\zeta$ function) works pretty well for similar problems, too.
answered Sep 17 '16 at 13:01
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$\begingroup$ Very cool, thank you. Can the same technique be used to find $\sum_{n = 1}^{\infty}[\zeta(mn) - 1]$ for an arbitrary positive integer $m$? I think it's interesting that these series of irrational numbers without any obvious pattern converge to such simple rational limits. $\endgroup$ – Vik78 Sep 17 '16 at 15:22
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$\begingroup$ @Vik78: you may use the same technique to compute the generating function $$f(x)=\sum_{n\geq 2}(\zeta(n)-1)x^{n} $$ in terms of the dilogarithm. If you apply a discrete Fourier transform to the above generating function (i.e. compute its values at the $m$-th roots of unity) you may compute such generalized series, too. They won't have rational values for every $m\in\mathbb{N}$, however. $\endgroup$ – Jack D'Aurizio Sep 17 '16 at 15:30
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The family $(n^{-k})$ with $n, k \ge 2$ is summable, because the geometric series gives us the estimation
$$\sum_{k=2}^N \sum_{n=2}^N \frac{1}{n^k} < \sum_{n=2}^N \sum_{k=2}^\infty \frac{1}{n^k} = \sum_{n=2}^N \frac{1}{n(n-1)} = 1 - \frac{1}{N} < 1.$$
Therefore we can exchange the order of summation and write
$$\sum_{k=2}^\infty (\zeta(k) - 1) = \sum_{n=2}^\infty \sum_{k=2}^\infty \frac{1}{n^k} = \sum_{n=2}^\infty \frac{1}{n(n-1)} = 1.$$
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Figured out the solution:$\sum_{n=2}^{\infty} [\zeta(n) - 1] = \sum_{n=2}^{\infty}\sum_{k=2}^{\infty} n^{-k} = \sum_{n=2}^{\infty}\frac{1}{1-\frac{1}{n}} - 1 - \frac{1}{n} = \sum_{n=2}^{\infty}\frac{1}{n(n-1)}$, which converges by comparison with $\zeta(2)$.
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$\begingroup$ The final sum converges to exactly $1$ (telescoping sum). $\endgroup$ – Erick Wong Sep 18 '16 at 19:36
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$$0\le \zeta(n)-1=\frac{1}{2^n}+\sum_{k\ge3}\frac{1}{k^n}\le\frac{1}{2^n}+\int_2^\infty\frac{dx}{x^n}=\frac{1}{2^n}+\frac{1}{(n-1)2^{n-1}}=\frac{n+1}{n-1}\frac{1}{2^n}$$
which $\le\frac{3}{2^n}$ for $n\ge2$. Thus, the sum converges by comparison with the geometric series $\sum \frac{1}{2^n}$.
