How can I calculate $\lim_{n \to \infty} (1 + \frac{1}{n!})^n$ and $\lim_{n \to \infty} (1 + \frac{1}{n!})^{n^n}$?

Question

How do you calculate the following limits?

$$\lim_{n \to \infty} \left(1 + \frac{1}{n!}\right)^n$$ $$\lim_{n \to \infty} \left(1 + \frac{1}{n!}\right)^{n^n}.$$

I really don't have any clue about how to proceed: I know the famous limit that defines $e$ ($\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n=e$), but the factorials (and the exponent of the second one) here throw me off. Any ideas?

Answer 1

HINT: $$ \lim_{n \to \infty} \left(1 + \frac{1}{n!}\right)^n=\lim_{n \to \infty} \left(\left(1 + \frac{1}{n!}\right)^{n!}\right)^{n/n!} $$ and the inner limit is $e$, hence the final one is 1. The second case is similar.

Answer 2

Consider the first expression $$A=\left(1 + \frac{1}{n!}\right)^n$$ So $$\log(A)=n\log\left(1 + \frac{1}{n!}\right)\approx \frac n{n!}=\frac 1{(n-1)!}$$ So, $\log(A)$ goes to $0$ and then $A$ to $1$.

Consider the second expression $$B_n=\left(1 + \frac{1}{n!}\right)^{n^n}$$ So $$\log(B)={n^n}\log\left(1 + \frac{1}{n!}\right)\approx \frac {n^n}{n!}$$ Now, use Stirling approximation $$n!\approx {n^n}\sqrt{2\pi n}e^{-n}$$ So, $$\log(B)\approx \frac {n^n}{{n^n}\sqrt{2\pi n}e^{-n}}=\frac {e^n}{\sqrt{2\pi n}}$$ So, $\log(B)$ goes to $\infty$ and then $B$ to $\infty$.

Answer 3

First limit is easy. Let $$f(n) = \left(1 + \frac{1}{n!}\right)^{n}$$ We will use the fact that the sequence $a_{n} = (1 + (1/n))^{n}$ is strictly increasing for $n \geq 3$ and tends to a positive limit (when $n \to \infty$) denoted by $e$. Also $2 < e < 3$. These facts can be proven (and given in many textbooks) without any standard theory of logarithms and exponentials. Another fact which we need to know is that if $A > 0$ then $A^{1/n}$ tends to $1$ as $n \to \infty$.

We thus have $a_{n} = (1 + (1/n))^{n} < e$ for all $n$. Replacing $n$ by $n!$ we get that $(1 + (1/n!))^{n!} < e$. Hence we can see that $$1 < f(n) = \left(1 + \frac{1}{n!}\right)^{n} = \left(\left(1 + \frac{1}{n!}\right)^{n!}\right)^{1/(n - 1)!} < e^{1/(n - 1)!}$$ Again it is easy to prove that $1/(n - 1)! < 1/n$ for all $n > 3$ and hence we have $$1 < f(n) < e^{1/n}$$ for $n > 3$. Letting $n \to \infty$ and using Squeeze theorem we get $f(n) \to 1$ as $n \to \infty$.

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For the function $$g(n) = \left(1 + \frac{1}{n!}\right)^{n^{n}}$$ we need to use the fact that $$a_{n} = \left(1 + \frac{1}{n}\right)^{n} \geq 2$$ for $n$. Replacing $n$ by $n!$ we get $$\left(1 + \frac{1}{n!}\right)^{n!} \geq 2$$ for all $n$. And then we can see that $$g(n) = \left(1 + \frac{1}{n!}\right)^{n^{n}} = \left(\left(1 + \frac{1}{n!}\right)^{n!}\right)^{n^{n}/n!}\geq 2^{n^{n}/n!}\tag{1}$$ We next need to analyze the sequence $n^{n} / n!$. It turns out it is simpler to handle its reciprocal $b_{n} = n!/n^{n}$. Clearly $$\frac{b_{n + 1}}{b_{n}} = \frac{(n + 1)!}{(n + 1)^{n + 1}}\cdot\frac{n^{n}}{n!} = \dfrac{1}{\left(1 + \dfrac{1}{n}\right)^{n}} \to \frac{1}{e} < 1$$ and hence the series $\sum b_{n}$ is convergent (Ratio test) and hence $b_{n} \to 0$.

It follows that $1/b_{n} \to \infty$ as $n \to \infty$. It also follows that $2^{1/b_{n}} \to \infty$ as $n \to \infty$. From $(1)$ we see that $g(n) \geq 2^{1/b_{n}}$ and hence $g(n) \to \infty$ as $n \to \infty$.

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Note: From the above it is seen that in most cases basic limit theorems and certain standard limits are sufficient to solve a problem without invoking advanced results. In this case for example we dealt with limits involving variable exponents like $n!$ and $n^{n}$ and yet their solution was possible without any theory of exponentials and logarithms. Unless we have to deal with irrational exponents (or say complex/imaginary) it does not make sense to invoke the general theory of logarithm and exponential.