Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $

This is one of the cutest applications of AM-GM I have learned. Unfortunately, I do not remember the source.

Define the numbers $x_0, x_1, x_2, \ldots, x_n$ by: $$ x_i = \begin{cases} 1, &i = 0, \\\\ 1+\frac{1}{n}, &1 \leqslant i \leqslant n. \end{cases} $$ The claim follows by applying AM-GM: $$ \left( \frac{x_0 + x_1 + \ldots + x_n}{n+1} \right)^{n+1} \gt \ \prod_{i=0}^n \, x_i . $$ Plugging in the above values, we get $$ \left( \frac{1+n \Big(1+\frac{1}{n} \Big)}{n+1} \right)^{n+1} \gt \ 1 \cdot \left( 1+\frac{1}{n} \right)^n , $$ which simplifies to $$ \left( 1+ \frac{1}{n+1} \right)^{n+1} \gt \left( 1 + \frac{1}{n} \right)^n. $$

Here's a direct argument without using AM-GM: write $$\left(1+{1\over n}\right)^n=\sum_{j\geq 0} {n\choose j}\left({1\over n}\right)^j=\sum_{j\geq 0}\,\, \prod_{0\leq k<j}\left(1-{k\over n}\right) \cdot{1\over j!}.$$ Each product inside the sum gets bigger as $n$ increases, and so the same is true for whole sum.

As requested, here is a proof using Bernoulli's inequality.

$(1+x)^r \ge 1 + rx$, for any real $x \gt -1$ and real $r \ge 1$.

We set $r = \frac{n+1}{n}$ and $x = \frac{1}{n+1}$.

We get

$$ \left(1 + \frac{1}{n+1}\right)^{(n+1)/n} \ge 1 + \frac{1}{n}$$

Taking $n^{th}$ power on both sides gives us the inequality.

$$ \left(1 + \frac{1}{n+1}\right)^{n+1} \ge \left(1 + \frac{1}{n}\right)^n$$

Now we only need to eliminate the equality portion.

Assume they were equal, then we must have that

$$(n+2)^{n+1}n^n = (n+1)^{2n+1}$$

which is not possible as $n+1$ is relatively prime with both $n$ and $n+2$. (Of course, we could probably have used a strict version of Bernoulli's inequality...).

The calculus argument: taking logarithms of $(1+1/n)^n$, it's enough to show that $f(x) = x \log (1+1/x)$ is an increasing function of $x$ for $x > 0$. Now $$ f^\prime(x) = \log \left( 1 + {1 \over x} \right) - {1 \over x+1} $$ and it suffices to show this is positive. So we need $\log (1 + 1/x) > 1/(x+1)$; taking exponentials it suffices to show that $1 + {1 \over x} > \exp \left( {1 \over x+1} \right)$ when $x > 0$. But we have $$ \exp(z) = 1 + z + {z^2 \over 2!} + {z^3 \over 3!} + \cdots < 1 + z + z^2 + z^3 + \cdots = {1 \over 1-z} $$ whenever $|z|<1$. Letting $z = 1/(x+1)$ gives $e^{1/(x+1)} < 1 + 1/x$, as desired.