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I have another series in mind, today it is $$\sum^\infty_{n=10}\sin\left(\frac{1}{n^3}+\frac{\cos(n)}{n^2}\right)$$ I have tried to investigate the argument: it is basically $\frac{1+n\cos(n)}{n^3}$ or we could say that $1/n^3$ converges, $1/n^2$ also and $\cos(n)$ have bounded partial sums, so the argument should converge. However, we are not talking about a sum of sines.

I tried also dividing the sum into two sums using summation rule for sine: $\sin(...)=\sin(1/n^3)\cos(\cos(n)/n^2)+\sin(\cos(n)/n^2)\cos(1/n^3)$, however, $\cos(1/n^3)$ does not converge, so it is not bounded and I am afraid that $\sin(\cos(n)/n^2)$ is not bounded either.

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  • $\begingroup$ Get used to thinking about LIMIT COMPARISON: What does the $n^{\text{th}}$ term "look like" in absolute value? If $|a_n| \sim c b_n$ for some $c\ne 0$ and you know that $\sum b_n$ converges, then the same is true for $a_n$. What does $\sim$ mean? For conditional convergence (or divergence), things are more delicate. $\endgroup$
    – Ted Shifrin
    Jun 22, 2014 at 14:51

1 Answer 1

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$|n^{-3}+n^{-2}\cos n|\le 2n^{-2}$, and $|\sin x|\le |x|$; hence the terms of the series are bounded in absolute value by the respective terms of the convergent series $\sum 2n^{-2}$, and the series converges absolutely.

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  • $\begingroup$ Oh right, forgot to test absolute convergence. LOL. $\endgroup$
    – user74200
    Jun 22, 2014 at 14:51

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