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I'm having trouble resolving the following limit: $$ \lim_{n \to \infty} \left(1+\frac{1}{n!}\right)^n $$ Intuituvely the limit is equal to 1, but the exercises requires me to resolve via calculation and I have no idea how I can accomplish this.
Can someone please explain it to me?

Sceptwo
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5 Answers5

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I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.

Using the inequality $1+x\le \frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $n\ge 2$

$$1\le \left(1+\frac1{n!}\right)^n\le \frac1{\left(1-\frac1{n!}\right)^n}\le \frac1{1-\frac1{(n-1)!}}$$

whereby application of the squeeze theorem yields the coveted limit.

Mark Viola
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    It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+x\le \frac1{1-x}$ as $(x<1)$ and take is as a given. – user Nov 29 '18 at 17:00
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    Well, for all $x$, $x^2\ge 0$. Hence, $(1+x)(1-x)=1-x^2\le 1$. For $x<1$, $1+x\le \frac1{1-x}$. So, that is a fairly standard "trick." – Mark Viola Nov 29 '18 at 17:04
  • Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :) – user Nov 29 '18 at 17:07
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    Yes, this "trick" can be used, for example, to show that $\exp(x)\exp(y)=\exp(x+y)$ where we use the limit characterization $\exp(x)=\lim_{n\to\infty}\left(1+\frac xn\right)^n$ of the exponential function. – Mark Viola Nov 29 '18 at 17:09
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HINT

We have

$$\lim_{n \to \infty} \left(1+\frac{1}{n!}\right)^n=\lim_{n \to \infty} \left[\left(1+\frac{1}{n!}\right)^{n!}\right]^{\frac1{(n-1)!}}$$

then refer to standard limit for $e$. How can we conclude form here?

user
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It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that $$ n\log\left(1+\frac{1}{n!}\right)=n\times \frac{1}{n!}\times \frac{\log\left(1+\frac{1}{n!}\right)}{1/n!} $$ Now let $n\to \infty$. Using the fact that $$ \lim_{u\to 0}\frac{\log(1+u)-\log1}{u-0}=1 $$ by definition of the derivative we get that $$ \frac{1}{(n-1)!}\times \frac{\log\left(1+\frac{1}{n!}\right)}{1/n!}\to0\times 1=0 $$ as $n\to \infty$.

Sri-Amirthan Theivendran
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Note that $$ \frac{1}{n!}<\frac{1}{n(n-1)}<\frac{1}{(n-1)^2} $$ Now prove that $$ \lim_{t\to0}\frac{\log(1+t^2)}{t}=0 $$ and observe that \begin{align} \left(1+\frac{1}{(n-1)^2}\right)^n&= \left(1+\frac{1}{(n-1)^2}\right)\left(1+\frac{1}{(n-1)^2}\right)^{n-1} \\[6px] &=\left(1+\frac{1}{(n-1)^2}\right)\exp\left(f\left(\frac{1}{n-1}\right)\right) \end{align} Finally note that $$ 1\le\left(1+\frac{1}{n!}\right)^n $$

egreg
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Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.

We use the following lemma of Thomas Andrews (proved using binomial theorem):

Lemma: If $\{a_n\} $ is a sequence of real or complex terms such that $n(a_n-1)\to 0$ then $a_n^n\to 1$.

Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!\to 0$. The desired limit is $1$ by the above lemma.

Paramanand Singh
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