# Does the series $f(n) = \sum_{n=0}^\infty e^{in} \frac{1}{\sqrt{n}}$ converge, and if so, does it converge slowly?

The series $$f(n) = \sum_{n=0}^\infty \frac{1}{\sqrt{n}},$$

doesn't converge as it is a p-series with $p = 1/2 <1$. If I add an oscillating term to the series such that

$$f(n) = \sum_{n=0}^\infty e^{in} \frac{1}{\sqrt{n}},$$

what can I say about this series? Does it now converge? If so, is it slow to converge? That is, would I need to take a large number of terms to get an accurate numerical approximation?

• you mean $\sum_{n=1}^\infty$ – reuns Oct 17 '16 at 12:23
• The series converges by Dirichlet's test. – achille hui Oct 17 '16 at 12:37
• @achillehui it is the same thing as the summation by parts – reuns Oct 17 '16 at 12:48

Use the summation by parts :$$\sum_{n=1}^N z^n n^{-s} = N^{-s} \sum_{n=1}^N z^n + \sum_{n=1}^{N-1} (n^{-s}-(n+1)^{-s}) \sum_{k=1}^n z^k$$ $$= N^{-s} \frac{z^{N+1}-1}{z-1}+ \sum_{n=1}^{N-1} (n^{-s}-(n+1)^{-s}) \frac{z^{n+1}-1}{z-1}$$ and the conclusion is that $n^{-s}-(n+1)^{-s} = \int_n^{n+1} s x^{-s-1}dx = sn^{-s-1}+ \mathcal{O}(n^{-s-2})$
The function $Li_s(z) = \sum_{n=1}^\infty z^n n^{-s}$ is the polylogarithm, and your series is $Li_{1/2}(e^i)$.
A somewhat lower-powered calculation uses that $$\frac{1}{\sqrt{n}}-\frac1{\sqrt{n+1}} =\frac1{\sqrt{n}\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})} \sim\frac1{2n^{3/2}}$$ so that the series on the right of $$(1-z)\sum_{n=1}^\infty\frac{z^n}{\sqrt{n}}=z-\sum_{n=2}^\infty\left(\frac{1}{\sqrt{n-1}}-\frac1{\sqrt{n}}\right)z^n$$ converges absolutely for all $z:|z|\le 1$.
• The same semi-telescoping trick works on the partial sums. As $z=e^i\ne 1$, the additional factor $(1-z)$ does not change the convergence behavior. – Lutz Lehmann Jan 7 '17 at 1:18
Starting from @user1952009's answer $$\sum_{n=1}^\infty \frac{e^{i n}}{\sqrt{n}}=\text{Li}_{\frac{1}{2}}\left(e^i\right)$$ we can expand the result in terms of the zeta function and get $$\sum_{n=1}^\infty \frac{e^{i n}}{\sqrt{n}}=\frac{\zeta \left(\frac{1}{2},\frac{1}{2 \pi }\right)+\zeta \left(\frac{1}{2},1-\frac{1}{2 \pi }\right) } 2+i\frac{\zeta \left(\frac{1}{2},\frac{1}{2 \pi }\right)-\zeta \left(\frac{1}{2},1-\frac{1}{2 \pi }\right)} 2$$