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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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    Nope, both of them taken from the web2012-03-20

2 Answers 2

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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.2012-03-20
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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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    @Ary$a$$b$hata I have edited the answer.2012-03-17