$\lim\limits_{n\to\infty}a^{n}\left(\frac{n+1}{n}\right)^{n^{2}}$ 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 remark: a fuction is said to converge or not, not a limit. If a function converges than the limit exists (is finite). Dec 14 '14 at 0:14
• This diverges for $a>1$. Can you see why? Now try when $0 < a < 1$. Dec 14 '14 at 0:16
• Hint: $\left(\frac{n+1}{n}\right)^n \to e$. Dec 14 '14 at 0:18
• @VincenzoOliva You don't mean "function" here, you mean "sequence". A sequence is said to converge. As a further remark, existence of a limit and being finite are two different things. $(-1)^n$ remains finite but doesn't converge.
– Joel
Dec 14 '14 at 0:57
• @Joel Well, yes. Apologies, two a.m. here, I had regarded it as a function. As for the other remark, I meant "not counting $\infty$". And I meant *then. Dec 14 '14 at 1:00

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 $a\in(-\frac{1}{e},\frac{1}{e}]$.

• But it doesn't converge for $a = -1/e$. Dec 14 '14 at 1:30
• Indeed. For $a=-1/e$ the sequence oscillates between $e^{-1/2}$ and $-e^{-1/2}$. Dec 14 '14 at 8:11
• @AntonioVargas You're right; I realized this on my way home last night. Thanks. Dec 14 '14 at 16:28

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 $a \in (-\frac 1 e, \frac 1 e]$.

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.
• When it is 0 multiply infinity, we cannot just treat them separately? Dec 14 '14 at 0:49
• What do you mean? $\infty \times 0$ is an indeterminate form, but if one of two factors tends to infinity, the only hope for convergence is that the other tends to 0. Can you see why? Dec 14 '14 at 0:52
• @MFSO_Zhou Maybe I clarified the thing, see the edit. Dec 14 '14 at 1:05
• I think you mean something else instead of $e^{-1}<a\le e^{-1}$. Dec 14 '14 at 19:57
• @user84413 Oops, thanks, good catch! Dec 14 '14 at 20:00

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 $-e^{-1/2}$