How to prove that sequence $(1+1/n)^n$ is convergent and increasing?

Question

For the sequence $(1+1/n)^n$, how does one prove that it is convergent and increasing series? I do know that as $n \to \infty$ it becomes constant $e$.

Answer 1

It is also true that $$ \left( 1 + \frac{1}{n} \right)^{n+1} $$ decreases. So, with $$ \left( 1 + \frac{1}{n} \right)^{n} < \left( 1 + \frac{1}{n} \right)^{n+1} $$ and getting arbitrarily close together, the two bang into each other somewhere. One approach uses $$ n \geq 2 \; \; \Longrightarrow \; \; \frac{1}{1+n} < \; \log \left( 1 + \frac{1}{n} \right) < \frac{1}{n} $$ The power series, with remainder, for $\log (1+x)$ actually tells us the slightly stronger $$ n \geq 2 \; \; \Longrightarrow \; \; \frac{1}{n} - \frac{1}{2n^2} < \; \log \left( 1 + \frac{1}{n} \right) < \frac{1}{n} $$

Answer 2

Here is an intuitive monetary argument. Imagine you are charging 100% interest. What do you want to do? Compound often. If you do $n$ compoundings in a period, your total balance at the end of one year is $$\left(1 + {1\over n}\right)^n.$$ More compoundings is more money. But there is a limit....

Answer 3

With elementary manipulations you can prove this sequence is limited from above by 3.

$$e_n = \left(1+\frac{1}{n}\right)^n = \sum_{k=0}^n{n \choose k}\frac{1}{n^k} = 1+1+\sum_{k=2}^n{n \choose k}\frac{1}{n^k} \leq $$ $$ 2 + \sum_{k=2}^n\frac{1}{k!} \leq 2+ \sum_{k=2}^n\frac{1}{2^{k-1}} \leq \sum_{k=2}^n 2+\sum_{k=2}^\infty \frac{1}{2^{k-1}} = 3 $$

Now if you prove that $e_n$ is monotonic increasing, the convergence follows from the existence of an upper bound.

You can find the answer of that step in this question: Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $

Answer 4

Whoops! I typed this out and forgot to submit it!

Let $a_n = (1+1/n)^n$. By the arithmetic-geometric mean inequality $$ a_n^{1/n} = 1+1/n \leq \frac{1+\sum_{k=1}^n(1+1/n)}{n+1} = \frac{n+2}{n+1} = 1+\frac{1}{n+1}, $$ and so (since $x \mapsto x^n$ is increasing and $1 + 1/(n+1) > 1$) $$ a_n \leq \left(1 + \frac{1}{n+1}\right)^n \leq \left(1 + \frac{1}{n+1}\right)^{n+1} = a_{n+1}. $$

This shows that $\{a_n\}$ is increasing. Hence we'll know the sequence converges if it is bounded above. There are many ways to see that this is true. For example, by the binomial theorem

$$a_n = \sum_{k=0}^n \binom n k n^{-k}$$

for all $n$, and the identity

$$ \binom n k \leq 2\left(\frac{n}{2}\right)^k $$

(which holds for all $n \geq 0$ and $0 \leq k \leq n$) implies

$$ a_n \leq 2\sum_{k=0}^n 2^{-k} = 2\frac{1-(1/2)^{n+1}}{1-1/2} = 4(1-(1/2)^{n+1}) < 4 $$ for all $n \geq 1$. Hence $\{a_n\}$ is bounded above.