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?

  • $\begingroup$ The terms aren't positive, so limit comparison doesn't just not work nicely. It can't be applied at all. $\endgroup$ – user296602 Mar 30 '16 at 3:56
  • $\begingroup$ $\sum s_nt_n$ converges if $\sum t_n$ form a bounded sequence and $s_{i+1}\le s_i$ with $s_n\rightarrow 0$. $\endgroup$ – vnd Mar 30 '16 at 4:01
  • $\begingroup$ 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. $\endgroup$ – 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?)

Edit: As @Michael pointed out, some properties of alternating series may help in the process of determining how many terms is needed.


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 $$\sum_{n=1}^\infty (-1)^n \sin \left( \frac{1}{n} \right)\approx -\log (2)+\frac{\zeta (3)}{8}-\frac{\zeta (5)}{128}+\frac{\zeta (7)}{5120}\approx -0.550794$$


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