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for what values of a does this series converge\diverge, absolutely converge diverge

$\sum_{n=1}^{\infty}{(\frac{an}{n +1})^n} , a\in \mathbb{R}$

at first i wanted to do the root test, but the problem is that $a$ could be negative, if i take it out of the fraction to get $(\frac{a}{n+1})^n$ it doesent meet the critiria of either Dirichlet, or Abel tests

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If $a\ne0$ then $$\left(\frac{an}{n+1}\right)^n=a^n \cdot \left(\frac{n}{n+1}\right)^n = a^n \cdot \left( \frac{n+1}{n} \right)^{-n} =a^n\left(1+\frac{1}{n}\right)^{-n}\sim e^{-1}a^n$$ so the given series is convergent if and only if $|a|<1$. The result is clear for $a=0$.

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  • $\begingroup$ can you explain the first thing you did? $\endgroup$ – guynaa Dec 7 '13 at 14:49
  • $\begingroup$ I used these formula $(ab)^n=a^nb^n$ and $a^n=(\frac{1}{a})^{-n}$ $\endgroup$ – user63181 Dec 7 '13 at 14:50
  • $\begingroup$ any change of adding a few steps? , i cant seem to achieve that result on my own $\endgroup$ – guynaa Dec 7 '13 at 15:01
  • $\begingroup$ $\sim\sim\sim +1$ $\endgroup$ – Namaste Dec 7 '13 at 15:21
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    $\begingroup$ @SamiBenRomdhane : I guess you mean $|a|<1$ when you write $|a|<0$... $\endgroup$ – user87543 Dec 7 '13 at 16:13
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Note that $\left(\frac n{n+1}\right)^n\to \frac1e$, so for $|a|\ge 1$ the summands do not even tend to $0$, whereas for $|a|<1$ you can compare with the geometric series.

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Hint: you can use the root test.

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  • $\begingroup$ you cant do that because $a$ might be negative $\endgroup$ – guynaa Dec 7 '13 at 15:02
  • $\begingroup$ @guynaa: offcourse you can use the root test! $\endgroup$ – Mhenni Benghorbal Dec 7 '13 at 15:07
  • $\begingroup$ @guynaa: if you apply this test you get $|ae^{-1}|<1$!! $\endgroup$ – Mhenni Benghorbal Dec 8 '13 at 3:48
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Hint:

$\sum a_n$ converges only if $a_n\rightarrow 0$

For $\sum (\frac{an}{1+n})^n$ Converge we need ???

Please see that $(1+\frac{1}{n})^n\rightarrow e$ ..

So, where should $a$ be sitting in?

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