For which values of $x$ is the following series convergent: $\sum_0^\infty \frac{1}{n^x}\arctan\Bigl(\bigl(\frac{x-4}{x-1}\bigr)^n\Bigr)$ For which values of $x$ is the following series convergent?
$$\sum_{n=1}^{\infty} \frac{1}{n^x}\arctan\Biggl(\biggl(\frac{x-4}{x-1}\biggr)^n\Biggr)$$

If $x>1$ the series converges absolutely because $\arctan$ is a bounded function:
$$
\Biggl|\frac{1}{n^x}\,\arctan\Bigl(\frac{x-4}{x-1}\Bigr)^n\Biggr|\le\frac{\pi}{2\,n^x}.
$$
If $x<1$ then $(x-4)/(x-1)>1$ and
$$
\lim_{n\to\infty}\arctan\Bigl(\frac{x-4}{x-1}\Bigr)^n=\frac{\pi}{2}.
$$
It follows that the series diverges in this case.
