# Prove that $\sum_{n=1}^{\infty} \sin \left(\frac{1}{n}+n\pi\right)$ converges

Prove that $\displaystyle \sum_{n=1}^{\infty} \sin \left(\frac{1}{n}+n\pi\right)$ converges and find its value to three decimal places.

I thought of using the limit comparison test but that doesn't work too nicely here. What other tests might work here? Also, how would I find its value to three decimal places and be sure of it?

• The terms aren't positive, so limit comparison doesn't just not work nicely. It can't be applied at all. – user296602 Mar 30 '16 at 3:56
• $\sum s_nt_n$ converges if $\sum t_n$ form a bounded sequence and $s_{i+1}\le s_i$ with $s_n\rightarrow 0$. – vnd Mar 30 '16 at 4:01
• If you sum 2000 terms you are within 0.0005 of the answer because it is alternating and all future terms are less than 0.0005. – Empy2 Mar 30 '16 at 4:05

Using the periodicity of $\sin$, $$\sin \left(\frac{1}{n} + n\pi \right) = (-1)^n \sin \left( \frac{1}{n} \right)$$ Then you can use the alternating series test. The numerical evaluation should be done using a calculator or computer. (Why?)
After Henry W's answer and just for your curiosity, we can approximate the sum using, to order $k$, the Taylor expansion of the sine. This gives $$\sin(\frac 1n)=\sum_{k=0}^p \frac{(-1)^k}{(2k+1)!} n^{-(2k+1)}+O\left(\frac{1}{n^{2p+3}}\right)$$ and take into account that, for $k>0$, $$\sum_{n=1}^\infty (-1)^n n^{-(2k+1)}=-(1-4^{-k})\,\zeta (2 k+1)$$ (for $k=0$,it reduces to $-\log(2)$.
Using $p=5$, the result is $$\sum_{n=1}^\infty (-1)^n \sin \left( \frac{1}{n} \right)\approx -\log (2)+\frac{\zeta (3)}{8}-\frac{\zeta (5)}{128}\approx -0.550991$$ while the infinite sum of the terms would be $\approx -0.550797$.
Using $p=7$, the result is $$\sum_{n=1}^\infty (-1)^n \sin \left( \frac{1}{n} \right)\approx -\log (2)+\frac{\zeta (3)}{8}-\frac{\zeta (5)}{128}+\frac{\zeta (7)}{5120}\approx -0.550794$$