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?
Sum of Reciprocals Raised to a Power
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sequences-and-series
analysis
limits
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3$$\sum_{n=1}^{ \infty} \frac{1}{n^x} = \zeta(x)$$ is finite forall $x \in (1, \infty)$, so the answer to your question is no. See [here](https://en.wikipedia.org/wiki/Riemann_zeta_function) for more details. – 2017-01-13
2 Answers
3
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 $s\le1$.