$\lim\limits_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}$ converges for $a$ in what range? 
$\displaystyle\lim_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}$
  converges for $a$ in what range?

I tried $\displaystyle\lim_{n\to\infty}\ln \left[a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}\right]=\lim_{n\to\infty}\left[n\ln a+n\ln\left(1+\frac{1}{n}\right)^{n}\right]=\lim_{n\to\infty}\left[n\ln(a)+n\right]$
$\displaystyle\Rightarrow a=\frac{1}{e}$, a particular one. But how can I get the range? 

You can use the root test for sequences:
Take the sequence in absolute value:
The limit of the sequence with an $n$'th root is $$L = \lim_{n \to \infty}\sqrt[n] {|a_n|} =\lim_{n \to \infty} |a| (1+\frac 1 n)^n = |a|e$$
The root test tells us that if $L \lt 1$ then the sequence converges (to $0$), and if $ L \gt 1$ it diverges. So we get that $|a_n|$ converges to $0$ for $|a| < \frac 1 e$, and if $|a_n|$ converges to $0$ it means that also $a_n$ converges (to $0$). A similiar argument tells us that $a_n$ diverges for $|a| > \frac 1 e$.
We are left to check $|a| = \frac 1 e$:
You've shown in the question that the sequence indeed converges when $a = \frac 1 e$. For $a = - \frac 1 e$ the sequence does not converge (Can be shown easily with taking 2 subsequences $a_{2n}, a_{2n + 1}$ and showing they have 2 different limits ($e^{1/2}$ and $-e^{1/2}$), and therefore the sequence diverges.
So in summary, we got $a_n$ converges iff $a \in (-\frac 1 e, \frac 1 e]$.
