Smallest $x$ for which $\sum_{n=1}^{\infty}\dfrac{1}{n^x}$ converges Consider the series
$$S_x = \sum_{n=1}^{\infty}\dfrac{1}{n^x}$$
for $x > 0$.
Then


*

*$S_1$ is the harmonic series, which is known to diverge.

*$S_2 = \dfrac{\pi^2}{6}$; this is the Basel problem solved by Euler

*$S_x$ appears to be well-studied or to have a famous history for a handful of other values of $x$ as well


My question is, is there a minimum known $x$ for which $S_x$ converges, or a maximum known value for which it diverges? And if the former, what is the sum $S_x$ for that minimum value of $x$?
 A: I'm positive this has been answered before, but it converges for all $x > 1$. This is easy to see with the integral test:
$$
\int_2^\infty t^{-x} \, dt = \left[ \frac{1}{1-x}t^{1-x} \right]_2^\infty < \infty
$$
You can use this to show that as $x \to 1$ the sum increases to infinity, too. Just bound the sum below by the integral.
A: Yes, you can upper-bound the series with the integral
$$I_p = \int_1^\infty x^{-p} dx$$
The integral will converge $\forall p > 1$, by simply integrating it like so:
$$
I_p = \int_1^\infty x^{-p} dx = \frac{x^{-1 - p}}{1 - p} \bigg |_1^\infty = \frac{1}{1 - p} \left ( \infty^{1 - p} - 1\right) 
$$
which will be equal to 
$$\frac{-1}{1 - p} = \frac{1}{p - 1}$$
if $p > 1$ since the first term of the evaluation, $\infty^{1 - p} = 0$  since $1 - p< 0$, hence the term becomes 0. (Note that the infinity business is not strictly rigourous, and I should actually take limits, but it's a good enough explanation)
If $p < 1$, then the term $\infty^{1 - p}$ is raised to some positive power, so the integral is divergent.
When $p = 1$, the integral becomes
$$I_1 = \int_1^\infty \frac{1}{x} dx = \ln x \bigg |_1^\infty = \ln \infty - \ln 1 = \text{divergent}
$$ 
Hence, $p = 1$ also diverges
