For what real values of $a$ does $\sum_{n=1}^{\infty}{(\frac{an}{n +1})^n}$ converge? 
For what real values of $a$ does the series
$$\sum_{n=1}^{\infty}\left(\frac{an}{n
+1}\right)^n$$
converge?

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 $$\left(\frac{a}{n+1}\right)^n$$ it doesn't meet the criteria of either the Dirichlet or the Abel tests.
 A: 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 $a=0$.
A: 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.
A: 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?
A: The question asks about real values of $a$, but, in order to make this answer more general (so that other versions of this question may be linked to this question as duplicates), I will assume that $\alpha \in \mathbb{C}$.  An expedient approach to the problem is to consider various tests of convergence.  Personally, I like to start with the simple ones, and work my way up.  To my mind, the simplest test of convergence is the ratio test:

Theorem (Ratio Test):  Let $(a_n)_{n=1}^{\infty}$ be a sequence of complex numbers, define
$$L = \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right|, $$
and let $S$ denote the infinite series $\sum_{n=1}^{\infty} a_n. $

*

*If $L < 1$, then the series $S$ is absolutely convergent (and, therefore, convergent).

*If $L > 1$, then the series $S$ is divergent.

*If $L = 1$, then the ratio test is inconclusive—the series $S$ may be divergent, conditionally convergent, or absolutely convergent.


In the case of the series in this question, $a_n = \left( \frac{an}{n+1} \right)^n$.  Thus
$$ L
= \lim_{n\to \infty} \left| \frac{ a^{n+1}(n+1)^{n+1} / (n+2)^{n+1} }{ a^n n^n / (n+1)^n } \right|
= \lim_{n\to\infty} \left| \frac{a(n+1)^{2n+1} }{ n^n(n+2)^{n+1} } \right|. \tag{1}$$
It can actually be shown that this limit is $|a|$, but the argument is a little tedious.  If we allow ourselves the use of big-$O$ notation,
$$ \frac{(n+1)^{2n+1}}{n^n(n+2)^{n+1}}
= \frac{n^{2n+1} + O(n^{2n})}{n^n(n^{n+1} + O(n^n))}
= \frac{n^{2n+1} + O(n^{2n}) }{n^{2n+1} + O(n^{2n})}
= \frac{n^{2n+1}}{n^{2n+1} + O(n^{2n})} + \frac{O(n^{2n}) }{n^{2n+1} + O(n^{2n})}. $$
As $n$ goes to infinity, the first term goes to $1$, and the second term vanishes.  Thus
$$ L
= \lim_{n\to\infty} \left| \frac{a(n+1)^{2n+1} }{ n^n(n+2)^{n+1} } \right|
= |a|.$$
That being said, this looks like the kind of problem which is often given to students in a first year calculus course (or, perhaps, a third year complex analysis course), which means that big-$O$ notation may not feel very rigorous.  A more convincing argument might be to compute the limit via binomial expansion of the terms in the fraction, but this is going to be quite tedious, as well.  Doable, but tedious.
Personally, I probably would have looked at the limit in (1) and decided to move on.  The next test I usually choose to apply is the root test.

Theorem (Root Test): Let $(a_n)_{n=1}^{\infty}$ be a sequence of complex numbers, define
$$ L = \lim_{n\to \infty} \sqrt[n]{|a_n|^{n}}, $$
and let $S$ denote the series $\sum_{n=1}^{\infty} a_n$.

*

*If $L < 1$, then the series $S$ is absolutely convergent (and, therefore, convergent).

*If $L > 1$, then the series $S$ is divergent.

*If $L = 1$, then the ratio test is inconclusive—the series $S$ may be divergent, conditionally convergent, or absolutely convergent.


With $a_n = \left( \frac{an}{n+1} \right)^n$, the limit of interest is
$$
L = \lim_{n\to\infty} \left| \left( \frac{a n}{n+1} \right)^n \right|^{1/n}
= |a| \lim_{n\to\infty} \left| \frac{n}{n+1} \right|
= |a|. $$
This is, I think, a much more straight-forward computation, but it gives us the same result as the ratio test; namely, the series of interest converges absolutely when $|a| < 1$, and diverges when $|a| > 1$.
On the other hand, when $|a| = 1$, both tests are inconclusive.  In general, there is not really a good way of checking for convergence on the boundary of the disc of convergence.  Indeed, there are examples of series which converge for some points on this boundary, and diverge for others (as a simple example, consider the harmonic series, as compared with the alternating harmonic series).  In this case, however, we can do a little better.  Observe that if $|a| = 1$, then
$$ \lim_{n\to \infty} \left|  \frac{an}{n+1} \right|^{n}
= \lim_{n\to\infty} \left[ |a|^n \left( 1 + \frac{1}{n} \right)^{-n} \right]
= \frac{1}{\mathrm{e}}.
$$
In other words, we know that when $|a| = 1$, the general term of the series fails to converge to $0$, which means that the series fails to converge (by the divergence test).  Therefore, we may conclude:

*

*If $|a| < 1$, then the series converges absolutely.

*If $|a| \ge 1$, then the series diverges.

