Compute $\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.$

Question

NOTE 1: L'Hospitals and Taylor expansions are not allowed.

NOTE 2: I really appreciate if someone would correct my attempt, however any other easier method only involving single variable calculus (excluding the concepts in NOTE 1) are welcome.

PROBLEM: Compute $$\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.$$

I'll just manipulate without writing out the limit, for now. I have

\begin{array}{lcl} \left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n} & = & \left( 1+\frac{1}{n}\right)^{n^2}\cdot e^{-n} \\ & = & \exp\left( n^2\ln\left(1+\frac{1}{n}\right)-n\right) \\ & = & \exp((n\ln(1+\frac{1}{n})-1)n) \\ \end{array}

And proceeding:

\begin{array}{lcl} \exp((n\ln(1+\frac{1}{n})-1)n) & = & e^{((n\ln(1+\frac{1}{n})-1)n} \\ & = & (e^{((n\ln(1+\frac{1}{n})-1)})^n \\ & = & \left(\frac{(e^{((n\ln(1+\frac{1}{n})-1)}-1+1)}{(n\ln(1+\frac{1}{n})-1}\cdot{((n\ln(1+\frac{1}{n})-1})\right)^n\\ \end{array}

It gets quite ugly very quickly as you can see. I'm trying to rewrite it so I can apply standard limits like $$\lim_{x\rightarrow\infty}\frac{e^x-1}{x}=\infty \quad \text{and} \quad \lim_{x\rightarrow\infty}\frac{\ln{(1+x)}}{x}=0.$$

Answer 1

$$\lim_{x\to 0}\frac{\log(1+x)-x}{x^2} = \lim_{x\to 0}\frac{1}{x^2}\int_{0}^{x}\frac{-t}{1+t}\,dt=\lim_{x\to 0}\int_{0}^{1}\frac{-t\,dt}{1+xt}\stackrel{\text{DCT}}{=}\int_{0}^{1}-t\,dt=\color{red}{-\frac{1}{2}} $$

hence the given limit equals $\color{red}{\large{\frac{1}{\sqrt{e}}}}$ by straightforward manipulations.
$\text{DCT}$ stands for the Dominated Convergence Theorem.

Answer 2

As mentioned in comments it is difficult to avoid advanced tools like L'Hospital's Rule or Taylor expansions here. Your approach leads us to the limit $$\lim_{x\to 0}\frac{\log(1+x)-x}{x^{2}}\tag{1}$$ (put $x=1/n$ in your approach and you get the above limit in exponent). A trivial application of Taylor or L'Hospital's Rule shows that the above limit is $-1/2$ so that the answer is $1/\sqrt{e}$.

Another idea is to put $1+x=e^{t}$ and the limit in equation $(1)$ gets transformed into $$-\lim_{t\to 0}\frac{e^{t}-1-t}{t^{2}}\tag{2}$$ Using $$e^{t} =\lim_{n\to\infty} \left(1+\frac{t}{n}\right)^{n}\tag{3}$$ and binomial theorem we can see that the above limit in $(2)$ is $1/2$. For details see this answer.

Answer 3

Using the link

"Find the smallest $a>0$ such that $(1+\frac{1}{x})^{x+a}>e$ for all $x\geq 1\,$."

we have $\enspace\displaystyle (1+\frac{1}{n})^n<e<(1+ \frac{1}{n} )^{n+\frac{1}{2}}\enspace$for positive $\,n\,$ and therefore

$\displaystyle \frac{1}{\sqrt{e}}< (1+\frac{1}{n})^{-\frac{n}{2}}< (1+ \frac{1}{n})^{n^2}e^{-n}<1\enspace$ .

But $\enspace\displaystyle (1+ \frac{1}{x} )^{x^2}e^{-x}\enspace$ is strictly monotonous decreasing for $\,x>0\,$:

One result in the answer for the question in the link above is $\enspace\displaystyle \ln(1+\frac{1}{x} )<\frac{x+\frac{1}{2}}{x(x+1)}\enspace$

with which we get $\enspace\displaystyle \frac{d}{dx}\left(x^2\ln(1+\frac{1}{x}) - x\right) = -\frac{2x+1}{x+1} + 2x\ln(1+\frac{1}{x})) <0 \enspace $ .

It follows $\enspace\displaystyle \lim\limits_{n\to\infty} (1+ \frac{1}{n})^{n^2}e^{-n} = \frac{1}{\sqrt{e}} \enspace$ .

Answer 4

Hint:

We take for granted that $e$ is the sum of the inverses of the factorials.

Then by the binomial theorem,

$$\left(1+\frac1n\right)^n=1+1+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}+\cdots\frac1{n^n}\\ =1+1+\frac12+\frac1{3!}+\cdots-\frac1{2n}-\frac{3n-2}{3!n^2}-\frac{6n^2-11n+6}{4!n^3}\cdots$$

The leading terms in the numerators of the fractions are of the form $\dfrac{(k+1)(k+2)}2n^k$, for denominators $n^{k+1}(k+2)!$, so that every fraction contributes a term $\dfrac1{2(k-1)!n}$ and other terms with higher powers of $n$.

So

$$\frac{\left(1+\dfrac1n\right)^n}e=1-\frac{1-t_n}{2n}+\frac{r_n}{n^2}=1+\frac{p_n}n $$ where $t_n$ is the tail of the summation of $e$ and $r_n$ is bounded above by a constant.

Now we have

$$\left(1+\frac{p_n}n\right)^n=\left(\left(1+\frac{p_n}n\right)^{n/p_n}\right)^{p_n},$$ which tends to $e^{\lim_{n\to\infty} p_n}=e^{-1/2}$.

---

This is tagged as a hint because the argument showing that $r_n$ is bounded is missing.

Answer 5

Let

$$a=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} \quad ;\quad b=\lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2}{e^n}$$

We've;

\begin{align} c= \frac ab =\frac{\displaystyle\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} }{\displaystyle\lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2} {e^n} } =\lim_{n \to \infty}\frac{\displaystyle \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} }{\displaystyle \left(1-\frac 1n\right)^{n^2} {e^n} } &=\lim_{n \to \infty}\left(\frac {n+1}{n-1}\right)^{n^2} \frac{1}{e^{2n}} \end{align}

Now, let $n=2m$. Since $n \to \infty$, $m \to \infty$ too.

\begin{align} c&= \lim_{n \to \infty}\left(\frac {n+1}{n-1}\right)^{n^2} \frac{1}{e^{2n}}\\ &=\lim_{m \to \infty}\left(\frac {2m+1}{2m-1}\right)^{(2m)^2} \frac{1}{e^{2(2m)}}\\ &=\lim_{m \to \infty}\left(1+\frac {2}{2m-1}\right)^{4m^2} \frac{1}{e^{4m}}\\ \end{align}

Let $m-\frac 12=p$.

\begin{align} c&=\lim_{m \to \infty}\left(1+\frac {2}{2m-1}\right)^{4m^2} \frac{1}{e^{4m}}\\ &=\lim_{p \to \infty}\left(1+\frac {1}{p}\right)^{4p^2+4p+1} \frac{1}{e^{4p+2}}\\ &=\lim_{p \to \infty}\left(1+\frac {1}{p}\right)^{4p^2} \left(1+\frac {1}{p}\right)^{4p}\left(1+\frac {1}{p}\right)\frac{1}{e^{4p} \cdot e^2}\\ &=\left(\left(1+\frac {1}{p}\right)^{p^2}\frac {1}{e^p}\right)^4 \left(1+\frac 1p\right)^{4p} \frac{1}{e^2}\\ &=a^4 \cdot e^4 \cdot \frac{1}{e^2}\\ &=a^4 e^2\\ \end{align}

Thus we have $\dfrac ab=a^4 e^2$ $$ \color{blue}{\implies a^3b=\frac{1}{e^2}} \tag 1$$

Now, we also have

\begin{align} ab=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} \cdot \lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2}{e^n}&=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n}\cdot \left(1-\frac 1n\right)^{n^2}{e^n}\\ &=\lim_{n \to \infty}\left(1-\frac {1}{n^2}\right)^{n^2}\\ &=\frac{1}{e} \end{align}

Hence,

$$\color{red}{\implies ab=\frac{1}{e}} \tag 2$$

Using $(1)$ and $(2)$ we finally have

$$\bbox[5px,border:2px solid #6b2fed]{a=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} =\frac{1}{\sqrt e}}$$

Answer 6

$\lim\limits_{n\to +\infty} a_ne^{n} =e^{-\frac{1}{2}}$. Setting $h=\frac{1}{n}$

$$a_n =\left(1+\dfrac{1}{n}\right)^{n^2} =\exp\left(\frac{\ln(1+\frac{1}{n})}{\frac{1}{n^2}}\right)=\exp\left(\frac{\ln(1+h)}{h^2}\right)$$ Thus,

$$\lim_{n\to +\infty} a_ne^{-n} =\lim_{h\to 0} = \exp\left(-\frac{1}{h}+\frac{\ln(1+h)}{h^2}\right)=e^{-\frac{1}{2}}$$

Given, that we know by Schwartz derivative that, if a function is $C^2$ near $x = 0$ we have,

taking $f(x) = \ln(1+x)$, $f(0)= 0$, $f'(0) =1$,$f''(0) =-1$

$$\color{red}{-\frac{1}{2} =\lim_{h\to 0} \frac{\frac{ \ln(1+h) }{h}-1}{h} =\lim_{h\to 0} -\frac{1 }{h}+\frac{ \ln(1+h) }{h^2}}$$

$$\color{red}{\frac{f''(0)}{2} =\lim_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}}$$ See here: How to prove Schwarz derivative $\frac{f''(0)}{2} =\lim_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}$ without Taylor expansion or L'Hopital rule?

Answer 7

(One idea only) I woul try to compute like this:

$$\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}=\underset{n\to\infty}{\lim}{\left(\frac{\displaystyle{\left(1+\frac{1}{n}\right)^{n}}}{e}\right)^{n}} = e^{\underset{n\to\infty}{\lim}{n\left(\frac{1}{e}\left(1+\frac{1}{n}\right)^{n}-1\right)}}=e^{-\frac{1}{2}} $$

and $\underset{n\to\infty}{\lim}{n\left(\frac{1}{e}\left(1+\frac{1}{n}\right)^{n}-1\right)}=-\frac{1}{2}$, but the only way I know to verify the last limit is by L'hôpital. Using the following theorem:

If $\underset{x\to a}{\lim}{f(x)}=1$ and $\underset{x\to a}{\lim}{g(x)}=\infty$, then we can show that: $$\underset{x\to a}{\lim}{f(x)^{g(x)}}=e^{\underset{x\to a}{\lim}{(f(x)-1)g(x)}}$$ without L'hôpital. I remember this from Demidovich book, but I don't know how to solve the limit $\underset{n\to\infty}{\lim}{n\left(\frac{1}{e}\left(1+\frac{1}{n}\right)^{n}-1\right)}=-\frac{1}{2}$ without l'hôpital.

Answer 8

As $n \rightarrow \infty$, both $\ln\left(1+\frac1n\right)$ and $\frac1n-\frac1{2n^2}$ tend to $0$ so

$$\ln\left(1+\frac1n\right)\rightarrow\frac1n-\frac1{2n^2}$$$$\Downarrow$$$$n^2\ln\left(1+\frac1n\right)\rightarrow n-\frac1{2}$$$$\Downarrow$$$$n^2\ln\left(1+\frac1n\right)-n\rightarrow -\frac12$$$$\Downarrow$$$$e^{n^2\ln\left(\frac{n+1}n\right)-n}\rightarrow e^{-\frac12}.$$Thus$$\lim_{n \rightarrow \infty}\left[\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}\right]=\frac1{\sqrt{e}}.$$