# for what $a$ values does this series converge?

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

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 $a=0$.

• can you explain the first thing you did? – guynaa Dec 7 '13 at 14:49
• I used these formula $(ab)^n=a^nb^n$ and $a^n=(\frac{1}{a})^{-n}$ – user63181 Dec 7 '13 at 14:50
• any change of adding a few steps? , i cant seem to achieve that result on my own – guynaa Dec 7 '13 at 15:01
• $\sim\sim\sim +1$ – Namaste Dec 7 '13 at 15:21
• @SamiBenRomdhane : I guess you mean $|a|<1$ when you write $|a|<0$... – user87543 Dec 7 '13 at 16:13

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.

Hint: you can use the root test.

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

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?