$\lim\limits_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}$ 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? 
 A: 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 $a\in(-\frac{1}{e},\frac{1}{e}]$.
A: 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 $a \in (-\frac 1 e, \frac 1 e]$.
A: Since $$\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^n=e,\tag{1} $$ we have $$\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^{n^2}=+\infty. $$Thus your function can only converge if $$\lim_{n\to\infty}a^n=0,$$ which implies $$|a|<1 \leftrightarrow -1<a<1$$ because that means multiplying by itself infinitely many times a number which can be written in the form of a fraction with the denominator larger than $1$. In our case, for some $b>1$, $$a^n=\left(\frac{1}{b}\right)^n=\frac{1}{b^n}.$$ Indeed, $$\lim_{n\to\infty}\frac{1}{b^n}=0.$$

More precisely, the sequence converges if and only if $a \in \left(-e^{-1}; e^{-1}\right],$ as long as $$\lim_{n\to\infty}e^{-n}\left(\frac{n+1}{n}\right)^{n^2}=\lim_{n\to\infty}e^{-n}\left(\left(\frac{n+1}{n}\right)^{n}\right)^n=l\ne0.\tag{2}$$ In fact, $l=e^{-1/2}$ but what really matters is that $(2)$ follows directly from $(1)$, and thus:


*

*for $a>e^{-1}$ the sequence is unbounded;

*for $-e^{-1}<a\le e^{-1}$ the sequence converges to $l\ne 0$;

*for $-e^{-1}<a<e^{-1}$ the sequence converges to $0$;

*for $a=- e^{-1}$ the sequence does not converge, oscillating between $l$ and $-l$;

*for $a<-e^{-1}$ the sequence is unbounded both below and above.

A: Let's suppose $a>0$. If the limit at infinity of 
$$
f(x)=\ln\Bigl(a^x\left(1+\frac{1}{x}\right)^{\!x^2}\Bigr)
$$
exists finite, then it's the same as the logarithm of the limit of the sequence; if the limit is $-\infty$, then the limit of the sequence is $0$; if the limit is $\infty$, then the limit of the sequence is $\infty$.
Using the function is handier; for instance, we can instead compute
$$
\lim_{t\to0^+}f(1/t)=
\lim_{t\to0^+}\left(\frac{\ln a}{t}+\frac{\ln(1+t)}{t^2}\right)=
\lim_{t\to0^+}\frac{t\ln a+\ln(1+t)}{t^2}=
\lim_{t\to0^+}\frac{(1+t)\ln a+1}{2t(1+t)}
$$
The numerator has limit $1+\ln a$; so if this is negative the limit is $-\infty$; if it is positive, the limit is $\infty$. If $1+\ln a=0$, we have
$$
\lim_{t\to0^+}\frac{-t}{2t(1+t)}=-\frac{1}{2}
$$
Thus we can conclude that your limit is
$$
\lim_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{\!n^{2}}=
\begin{cases}
0 & \text{if $0<a<1/e$} \\[6px]
e^{-1/2} & \text{if $a=1/e$} \\[6px]
\infty & \text{if $a>1/e$}
\end{cases}
$$
What if $a<0$? Well, for $a<-1/e$ the absolute value diverges; for $-1/e<a<0$, the absolute value converges to $0$, so also the sequence does.
For $a=0$ the sequence is constant.
It remains to see what happens for $a=-1/e$; this is easy, because the even terms converge to $e^{-1/2}$, while the odd terms converge to $-e^{-1/2}$
