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Help me please with these 2 questions:

1.Does it converge or diverge? : $$ \sum_{n=2}^{\infty }2^{n}\left ( \frac{n}{n+1} \right )^{n^{2}} $$

2.Check out absolute and conditional convergence of: $x>0 $

$$ \sum_{n=1}^{\infty }\sin (n)\sin \frac{x}{n} $$

Thanks a lot!

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Part 2 seems to be one of the harder homework type questions. Where did you find part 2? Is this homework? If so, has the class considered some previous problems with terms involving $\sin(n)$? – Aryabhata Mar 16 '12 at 20:31
Nope, both of them taken from the web – Lilly Mar 20 '12 at 8:24
up vote 7 down vote accepted

Hint for 1:

For sufficiently large $n$, $(\frac{n}{n+1})^n = (1 - \frac{1}{n+1})^n \le c$ for some $ 0 \lt c \lt \frac{1}{2}$. Why?

Now trying using the above to prove that your series converges.

For part 2, I believe you can use the Dirichlet Test to prove convergence.

To show that the series does not converge absolutely, use $\sin (x/n) \ge x/2n$ for sufficiently large $n$ and use the fact that at least one of $n$, $n+1$ is more than $\frac{1}{2}$ away from the multiple of $\pi$ which is closest to them.

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Thank you, it's clear now. – Lilly Mar 20 '12 at 8:23

Part 2

As indicated in Aryabatha's answer, convergence of $\sum(\sin n)\sin(x/n)$ follows from Dirichlet's test: $\sin(x/n)$ is eventually decreasing, converges to $0$ and the partial sums $\sum_{k=1}^n\sin k$ are bounded. To show that it does not converge absolutely, use the inequalities $$ \sin x\ge \frac{2\,x}\pi,\quad|\sin n|\ge\sin^2n=\frac{1-\cos(2\,n)}{2}. $$ Then $$ |(\sin n)\sin\Bigl(\frac{x}{n}\Bigr)|\ge\frac{x}{\pi}\Bigl(\frac1n-\frac{\cos(2\,n)}{n}\Bigr). $$ Again by Dirichlet'e test $\sum_{n=1}^\infty\cos(2\,n)/n$ converges. In particular there exists a constant $A>0$ such that $\Bigl|\sum_{k=1}^n\cos(2\,n)/n\Bigr|\le A$ for all $n$. Then $$ \sum_{k=1}^n|(\sin k)\sin\Bigl(\dfrac{x}{k}\Bigr)|\ge\frac{x}{\pi}\Bigl(\sum_{k=1}^n\frac1k-\sum_{k=1}^n\frac{\cos(2\,n)}{n}\Bigr)\ge\frac{x}{\pi}(\log(n+1)-A). $$

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Can you elaborate a bit more? Especially the last step... – Aryabhata Mar 17 '12 at 0:11
@Aryabhata I have edited the answer. – Julián Aguirre Mar 17 '12 at 7:15
Thanks........... – Aryabhata Mar 17 '12 at 17:11
Thanks, very good explanation! – Lilly Mar 20 '12 at 8:26

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