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I was asked to determine if the next sums converge absolutely, converge conditionaly or diverge.

enter image description here

For the first question I tried to use Leibniz:

Define $a_n=\frac{1}{n^a ln(n)}$. It's easy to show that $a_n$ is decreasing.

I found that if $a \geq 1$, then $a_n$ not converges (what does it say on the sum?) and if a<1, then $a_n$ converges to 0 and then the sum converges.

Can you help me continue and give hints for the other two.

Thanks!

Galc127
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1 Answers1

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$1)$ $$\frac {1}{({\frac {n+1}{an}})^n}=\frac {a^n}{({\frac {n+1}{n}})^n}$$

$2)$

$$ |(-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}\sin(\frac {1}{\sqrt n})|\leq |\frac {1}{\frac {(-1)^n+\sqrt n}{\sqrt n}}|$$

Haha
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  • Hi, I tried to continue according to your hints, but didn't succeed... may you show how to continue 1? I guess I'll be able to solve 2 after that. Thank you. – Galc127 Dec 09 '13 at 11:09
  • $(\frac {n+1}{n})^n\to e$. Now if $a>1=>a^n\to \infty$ and the series doesn't converge. Now,if $a=1$ the inside seqeunce goes to $1/e\neq 0$ so again ,diverges. If now $a<1=>a^n\to 0$.After that you can use the root test. – Haha Dec 09 '13 at 11:25
  • For the first hint, after a while $\left(\frac{n+1}{n}\right)^n$ is near $e$. The second hint as it is currently is not useful. – André Nicolas Dec 09 '13 at 11:26
  • May someone help with the second sum? still didn't manage to solve it. – Galc127 Dec 13 '13 at 08:40
  • $\left(\frac{n+1}{n}\right)^n\to e$ and thus we must have $a<1$. – Haha Dec 13 '13 at 09:36