Prove limit converges in definition of $e.$

Question

I've looked up several related questions, but they do not answer what I am looking for. Please give link if this is a duplicate.

What I eventually want to know is why

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n$$

converges to a finite value.

I mean, we have to know that the limit converges before we can call it a constant, right?

I know that $f(n)=\left(1+\dfrac1n\right)^n$ is strictly increasing, but how do we know it is bounded above?

Also, I tried using the binomial theorem to write

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left[\sum_{k=0}^n\frac{n!}{n^k k! (n-k)!}\right],\tag{1}$$

but I am unsure how to prove that the sum converges for $n\to\infty,$ because the general term depends on both $n$ and $k.$

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Here is what I think will work.

Let $a_{n,k}=\dfrac{n!}{n^k k! (n-k)!}.$ Then $a_{n+1,k+1}=\dfrac{(n+1)!}{(n+1)^{k+1} (k+1)! (n-k)!}.$

Using the Ratio Test, we get

$$\lim_{n\to\infty\atop k\to\infty}\left|\frac{a_{n+1,k+1}}{a_{n,k}}\right|=\lim_{n\to\infty\atop k\to\infty}\left[\frac{1}{k+1}\left(\frac{n}{n+1}\right)^k\right].$$

where the limit on the RHS is less than 1.

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That may take care of the case where $n\in\mathbb{N},$ so I suppose that (1) can be generalized using "Newton's generalised binomial theorem."

Answer 1

$$\begin{align}\left(1+\frac 1n\right)^n&=\sum_{k=0}^{n}\binom{n}{k}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots(n-k+1)}{k!}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{1}{k!}\left(1-\frac 1n\right)\left(1-\frac 2n\right)\cdots\left(1-\frac{k-1}{n}\right)\\&\lt \sum_{k=0}^{n}\frac{1}{k!}\\&=1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2}\\&\lt 1+\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\\&=3-\frac{1}{2^{n-1}}\\&\lt 3\end{align}$$

Answer 2

First you should know that: $$ (1) \ln(1+x)=x-\frac{x^2}{2}+O(x^3)$$ Also consider that ln is an injective function meaning if $\ln(A)=1$ then $A=e$
suppose $A=(1+\frac{1}{n})^n$ So: $$ \lim_{n\to\infty} \ln(A) = \lim_{n\to\infty} \ln((1+\frac{1}{n})^n = \lim_{n\to\infty} n\ln(1+\frac{1}{n}) $$ Applying (1): $$ \lim_{n\to\infty} n\ln(1+\frac{1}{n}) = \lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3}))$$ $$\lim_{n\to\infty} n(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3})) = \lim_{n\to\infty} (1-\frac{1}{2n}+nO(\frac{1}{n^3})) = 1 $$ Since $$ \lim_{n\to\infty} \ln(A) = 1 $$ Than $$ \lim_{n\to\infty} A = e $$ So e is the least upper bound.

Answer 3

I do not understand how your approach should work. First, we need to show that $\sum_{k=0}^n a_{n,k}$ is bounded. How the fact that the double limit $\lim_{n\to\infty\atop k\to\infty}\frac{a_{n+1,k+1}}{a_{n,k}}$ is smaller than $1$ (if it exists) is helpful for this? Next, $f(n)$ is strictly increasing, so for any positive $x$, $f(x)$ is bounded from above by $f(n)$ for any natural $n>x$, so why do we need to generalise binomial theorem for noninteger $n$?

On the other hand,

$$\sum_{k=0}^n\frac{n!}{n^k k! (n-k)!}\le \sum_{k=0}^n\frac{n!}{n^k(n-k)!}\cdot \frac 1{k!}\le$$ $$ \sum_{k=0}^\infty \frac 1{k!}\le 2+\sum_{k=2}^\infty \frac 1{k(k-1)} =2+\sum_{k=2}^\infty \left(\frac 1{k-1}-\frac 1{k}\right)=3.$$