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Recently I discovered that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{π^2}{6}$, and we know that sum of reciprocals of naturals to the first power diverges to infinity. So I was wondering just out of curiosity, whether there was a number between $1$ and $2$ where this sum of reciprocals raised to that power started diverging. Is there such a number?

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    $\begingroup$ $$\sum_{n=1}^{ \infty} \frac{1}{n^x} = \zeta(x)$$ is finite forall $x \in (1, \infty)$, so the answer to your question is no. See here for more details. $\endgroup$
    – Crostul
    Commented Jan 13, 2017 at 15:58

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This is what you need. https://en.wikipedia.org/wiki/Harmonic_series_(mathematics), find 6.3 - p-series

To sum up what you ask

If $k>1$ then $\sum_{n=1}^\infty \frac{1}{n^k}$ converges

If $k\leq 1$ then $\sum_{n=1}^\infty \frac{1}{n^k}$ diverges

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We have the integral test, which has

$$\int_1^\infty\frac1{x^s}\ dx\sum_{n=1}^\infty\frac1{n^s}<1+\int_1^\infty\frac1{x^s}\ dx$$

Which quickly shows it converges only if $s>1$ and diverges for $s\le1$.

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