How to prove that sequence $(1+1/n)^n$ is convergent and increasing? 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 $e$.
 A: 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 
$$  n \geq 2 \; \; \Longrightarrow \; \; \frac{1}{n} - \frac{1}{2n^2} < \; \log \left( 1 + \frac{1}{n} \right) <  \frac{1}{n}   $$
A: 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....
A: 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} $
A: 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.
