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

  • $\begingroup$ A remark: a fuction is said to converge or not, not a limit. If a function converges than the limit exists (is finite). $\endgroup$ Dec 14 '14 at 0:14
  • $\begingroup$ This diverges for $a>1$. Can you see why? Now try when $0 < a < 1$. $\endgroup$
    – Hubble
    Dec 14 '14 at 0:16
  • $\begingroup$ Hint: $\left(\frac{n+1}{n}\right)^n \to e$. $\endgroup$
    – Arthur
    Dec 14 '14 at 0:18
  • $\begingroup$ @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. $\endgroup$
    – Joel
    Dec 14 '14 at 0:57
  • $\begingroup$ @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. $\endgroup$ 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 $a\in(-\frac{1}{e},\frac{1}{e}]$.

  • $\begingroup$ But it doesn't converge for $a = -1/e$. $\endgroup$ Dec 14 '14 at 1:30
  • $\begingroup$ Indeed. For $a=-1/e$ the sequence oscillates between $e^{-1/2}$ and $-e^{-1/2}$. $\endgroup$ Dec 14 '14 at 8:11
  • $\begingroup$ @AntonioVargas You're right; I realized this on my way home last night. Thanks. $\endgroup$
    – user84413
    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.
  • $\begingroup$ When it is 0 multiply infinity, we cannot just treat them separately? $\endgroup$
    – MFSO1991
    Dec 14 '14 at 0:49
  • $\begingroup$ 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? $\endgroup$ Dec 14 '14 at 0:52
  • $\begingroup$ @MFSO_Zhou Maybe I clarified the thing, see the edit. $\endgroup$ Dec 14 '14 at 1:05
  • $\begingroup$ I think you mean something else instead of $e^{-1}<a\le e^{-1}$. $\endgroup$
    – user84413
    Dec 14 '14 at 19:57
  • $\begingroup$ @user84413 Oops, thanks, good catch! $\endgroup$ 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 $-e^{-1/2}$


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