Is the infinite series $\sum_n n^{i\theta}$ convergent for any values of $ \theta$?

Question

Is the infinite series $\sum_n n^{i\theta}$ convergent for any values of $\theta$?

I have seen tricks for converting $\sum \sin(n) $ into a telescoping series, but am stumped because $ n^{i\theta} = e^{i \theta \ln n} = \cos (\theta \ln n) + i \sin (\theta \ln n) $ doesn’t seem like could succumb to similar trick.

Any ideas? Thanks!

Answer 1

For it to converge, both real part and imaginary part should converge to some values, respectively. It seems like for all value of $theta$, neither is converging.

For a $\sum \cos(\theta \ln n)$ to converge, a necessary condition is $\cos(\theta \ln n)$ has $0$ as its limit which it apparently doesn't even when $\theta=0$.

Answer 2

if $\theta \in \mathbb R$, $|n^{i\theta}| = 1 \not\rightarrow 0$. This proves that the serie diverges

Answer 3

The abscissa of convergence for the series for the Riemann Zeta function, $\sum_{n=1}^{\infty} n^{-s}$, is $\sigma = 1$, so the series does not converge for any value with $\sigma <1$.