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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 $$\lim_{x\rightarrow\infty}\frac{e^x-1}{x}=\infty \quad \text{and} \quad \lim_{x\rightarrow\infty}\frac{\ln{(1+x)}}{x}=0.$$

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marked as duplicate by Guy Fsone, Xam, kingW3, Stefan4024, user99914 Nov 11 '17 at 21:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I think your final expression doesn't give an answer $\dfrac{e^*-1+1}{-}$ $\endgroup$ – Nosrati Oct 19 '17 at 10:11
  • $\begingroup$ It doesn't. I'm stuck there. It becomes something raised to infinite power. $\endgroup$ – Parseval Oct 19 '17 at 10:14
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    $\begingroup$ Good luck to you, Parseval! $\endgroup$ – Michael Rozenberg Oct 19 '17 at 10:15
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    $\begingroup$ @MichaelRozenberg - Hehe, yes I really need it. $\endgroup$ – Parseval Oct 19 '17 at 10:16
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    $\begingroup$ If we use the formula $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}=e^{x}$$ then it is possible to avoid L'Hospital's Rule and Taylor expansions. BTW the use of Taylor expansions (or something equivalent like definition of log as an integral) is the more natural approach for this problem. $\endgroup$ – Paramanand Singh Oct 19 '17 at 10:54
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$$\lim_{x\to 0}\frac{\log(1+x)-x}{x^2} = \lim_{x\to 0}\frac{1}{x^2}\int_{0}^{x}\frac{-t}{1+t}\,dt=\lim_{x\to 0}\int_{0}^{1}\frac{-t\,dt}{1+xt}\stackrel{\text{DCT}}{=}\int_{0}^{1}-t\,dt=\color{red}{-\frac{1}{2}} $$

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

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  • $\begingroup$ Perfect! (+1) :-) $\endgroup$ – user90369 Oct 19 '17 at 16:54
  • $\begingroup$ This was new for me +1. It appears that some overkills are really interesting and this is one of them. $\endgroup$ – Paramanand Singh Oct 19 '17 at 17:16
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    $\begingroup$ DCT really isn't necessary for these sorts of things. Simply notice that$$\frac t{1+x}\le\frac t{1+xt}\le t,\quad x>0,t\in[0,1]$$Similarly for $x<0$, but flipped. We may integrate easily and take the limit with squeeze theorem. $\endgroup$ – Simply Beautiful Art Oct 19 '17 at 23:40
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    $\begingroup$ @SimplyBeautifulArt: of course you are right, the above proof turns into a completely elementary one through your remark. On the other hand, it becomes $\approx 0.5\,cm$ longer, while I love one-liners :D $\endgroup$ – Jack D'Aurizio Oct 19 '17 at 23:43
  • $\begingroup$ Lol, I guess we all enjoy those one-liners. $\endgroup$ – Simply Beautiful Art Oct 19 '17 at 23:44
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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 $\enspace\displaystyle \lim\limits_{n\to\infty} (1+ \frac{1}{n})^{n^2}e^{-n} = \frac{1}{\sqrt{e}} \enspace$ .

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  • $\begingroup$ When we use derivatives, we may well use Taylor and L'Hospital's Rule. The primary point of avoiding these techniques is to see how far one can go by using just algebra. $\endgroup$ – Paramanand Singh Oct 19 '17 at 13:58
  • $\begingroup$ @ParamanandSingh : That's a wide interpretation of the conditions. But: If the OP says, my argumentation is not suitable, I will delete my answer, no problem. I hope my calculation gives the OP at least a new point of view and are therefore helpful. :-) $\endgroup$ – user90369 Oct 19 '17 at 16:48
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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.

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  • $\begingroup$ I'm not sure I understand the last bit. How can I apply the binomal theorem here? $\endgroup$ – Parseval Oct 19 '17 at 11:36
  • $\begingroup$ @Parseval: apply binomial theorem on $(1+(t/ n)) ^{n} $ and take limit. Some detailed analysis is necessary and it is available in an old answer of mine which I am searching. $\endgroup$ – Paramanand Singh Oct 19 '17 at 13:37
  • $\begingroup$ @Parseval : I have given the details in a linked answer. Do have a look. $\endgroup$ – Paramanand Singh Oct 19 '17 at 13:48
  • $\begingroup$ (+1) Nice solution, but I have a nice overkill too :D $\endgroup$ – Jack D'Aurizio Oct 19 '17 at 16:39
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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.

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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

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

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    $\begingroup$ You assume that $a, b$ exist. Moreover the step where you get $a^{4}$ requires justification. If we assume existence there is a reasonably easier proof. $\endgroup$ – Paramanand Singh Oct 19 '17 at 20:17
  • $\begingroup$ @ParamanandSingh I just realised a mistake in my solution. It'll take me 5-10min to fix. So please wait. After my edit, it'll be a complete rigorous proof. $\endgroup$ – Jaideep Khare Oct 19 '17 at 20:18
  • $\begingroup$ Also $ab=1/e$ implies $aba^{2}=1$ ie $a=\sqrt{e} $. $\endgroup$ – Paramanand Singh Oct 19 '17 at 20:20
  • $\begingroup$ @ParamanandSingh Everything is fixed now. $\endgroup$ – Jaideep Khare Oct 19 '17 at 20:27
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    $\begingroup$ You still assume that $a, b$ exist. That's a key assumption here on which you build everything else. By the way you also realize your mistake in previous version which was based on replacing $m-1/2$ by $m$. One should be wary of such replacements. $\endgroup$ – Paramanand Singh Oct 19 '17 at 20:29
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$\lim\limits_{n\to +\infty} a_ne^{n} =e^{-\frac{1}{2}}$. 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?

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(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.

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

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