limit of $\left(1+\frac{1}{n!}\right)^n$

Question

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?

Answer 1

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?

Answer 2

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.

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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.

Answer 3

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 $n\to \infty$.

Answer 4

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

Answer 5

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.