# Prove $(1 + \frac{1}{n})^n$ is bounded above

I've checked similar questions on the site but couldn't find satisfactory solutions or hints.

Also, is there a more general approach to proving whether a given sequence is bounded below or above?

• Well you could consider it as a limit n goes to infinity. It is a well known constant – imranfat Mar 11 '15 at 18:17

Hint: Use the binomial formula:

$$\left(1 + \frac{1}{n}\right)^n =\sum_{j=0}^n\frac1{j!}\frac{n!}{(n-j)!}\frac1{n^j}= 1+1+\sum_{j=2}^n\frac1{j!} \prod_{k=1}^{j-1}(1-k/n) < \ldots$$

By the binomial formula: \begin{eqnarray*} \left(1+\frac{1}{n}\right)^n=1+1+\sum_{k=2}^n\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k. \end{eqnarray*}

Notice that \begin{eqnarray*} \binom{n}{k}\cdot\left(\frac{1}{n}\right)^k=\frac{n(n-1)\cdots(n-k+1)}{n^k}\cdot\frac{1}{k!}<\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}. \end{eqnarray*}

So we get \begin{eqnarray*} \left(1+\frac{1}{n}\right)^n<2+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=3-\frac{1}{n}<3. \end{eqnarray*}

• Can't help but feel that @RRL left something for the OP to do but you've completely given them the answer to what is almost certainly a homework problem. Very nice complete answer though! – CameronJWhitehead Mar 11 '15 at 18:33
• @CameronJWhitehead - I agree with you that I was given the answer completely. I have ignored this and am continuing to work with RRL's hint. However, this is not a homework problem, I'm self-studying Apostol's Calculus after a very long break from studies. – Nishant S Mar 11 '15 at 18:41

My question here (What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?) has an elementary proof that, if $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$ then the $a_n$ are increasing and the $b_n$ are decreasing. Since $a_n < b_n$, all the $a_n$ are less than any of the $b_n$.

Since $b_5 = 46656/15625 < 3$, all the $a_n < 3$.

Hint: It is pretty easy to prove, and we'll known, that the limit is $e$. It is also easy to prove that a convergent sequence is bounded...