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

Question

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?

Answer 1

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*}

Answer 2

Hint: Use the binomial formula:

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

Answer 3

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 $a_n < 3$.

Answer 4

Hint: It is well-known that the limit is $e $. It is also well-known that a convergent sequence is bounded...

I've posted this on this site before, but couldn't find it. So here goes again: using a single interval upper and lower Riemann sum for $\ln x:=\int_1^x\dfrac 1t\operatorname dt$, we get $\dfrac x{n+x}\le\ln(1+\frac xn)\le\dfrac xn$.

So, $e^{\frac x{n+x}}\le 1+\dfrac xn\le e^{\frac xn}$.

Thus $e^{\frac {nx}{n+x}}\le (1+\dfrac xn)^n\le e^x$.

Now let $n\to \infty$, for the proof of convergence.

I believe I first saw this proof in Best and Penner's Calculus.

Answer 5

Just for the heck of it.

If $x > 0$ then

$\begin{array}\\ \ln(1+x) &=\int_0^x \dfrac{dt}{1+t}\\ &\lt x\\ \text{so}\\ n\ln(1+\frac1{n}) &\lt n(\frac1{n})\\ &=1\\ \text{so}\\ (1+\frac1{n})^n &\lt e\\ \end{array} $