$\lim\limits_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}$ 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? 

Let $c_n=a^n\left(1+\frac{1}{n}\right)^{n^2}$ and let $t=\frac{1}{a}$, so $\ln c_n=n\ln a+n^2\ln(1+\frac{1}{n})=n^2\ln(1+\frac{1}{n})-n\ln t$
where $\frac{1}{n}-\frac{1}{2n^2}<\ln(1+\frac{1}{n})<\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}\implies n-\frac{1}{2}<n^2\ln(1+\frac{1}{n})<n-\frac{1}{2}+\frac{1}{3n}$
$\implies (1-\ln t)n-\frac{1}{2}<n^2\ln(1+\frac{1}{n})-n\ln t<(1-\ln t)n-\frac{1}{2}+\frac{1}{3n}$,
$\hspace{.54 in}$ so $(1-\ln t)n-\frac{1}{2}<\ln c_n<(1-\ln t)n-\frac{1}{2}+\frac{1}{3n}$.
1) If $a>\frac{1}{e}$, then $t<e\implies(1-\ln t)n-\frac{1}{2}\to\infty\implies\ln c_n\to\infty\implies c_n\to\infty$.
2) If $a=\frac{1}{e}$, then $\ln c_n\to-\frac{1}{2}\implies c_n\to e^{-\frac{1}{2}}$.
3) If $0<a<\frac{1}{e}$, then $t>e\implies 1-\ln t<0\implies\ln c_n\to-\infty\implies c_n\to 0$.
Taking absolute values, we get that the sequence converges iff $a\in(-\frac{1}{e},\frac{1}{e}]$.
