Smallest $x$ for which $\sum_{n=1}^{\infty}\dfrac{1}{n^x}$ converges

Question

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 $x$?

Answer 1

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.

Answer 2

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