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

How could we prove that this inequality holds

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

where $n \in \mathbb{N}$, I think we could use the AM-GM inequality for this but not getting how?

-
Why not try taking derivative for $f(x) = \displaystyle (1 + \frac{1}{x})^x$? – Shuhao Cao Sep 15 '11 at 19:25
– Aryabhata Sep 15 '11 at 20:50
@Aryabhata That's also a beautiful solution; why don't you post it here as well, modifying it to this problem? (By the way, are you like a fan of Bernoulli's inequality? :)) – Srivatsan Sep 15 '11 at 20:54
@Srivatsan: I thought this question was mainly about using AM-GM, hence chose to comment. Yeah, Bernoulli's inequality is pretty neat :-) Simple and powerful, similar to AM-GM. – Aryabhata Sep 15 '11 at 21:20

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

-
For this, you do not need the full AM-GM inequality - only the version with n of n+1 values the same and this can be proved from Bernoulli's inequality. – marty cohen Sep 15 '11 at 23:04
@marty I haven't seen the weaker version of AM-GM. But what you are suggesting could be similar to Aryabhata's solution linked to in one of his comments. – Srivatsan Sep 15 '11 at 23:07

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.

-
+1, and it's fun to see something other than the AM-GM inequality sometimes. :) – Mike Spivey Sep 15 '11 at 20:06
@Mike Sorry! You beat me to the "binomial argument", but I decided to leave my solution up anyways. When I returned from class, your answer was gone. – Byron Schmuland Sep 15 '11 at 21:19
No worries. I thought your argument was cleaner, so I deleted my answer. – Mike Spivey Sep 15 '11 at 22:20

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...).

-
In general we could also prove this inequality (dirrectly) by using the inequality $(1+ x/a)^a \gt (1+x/b)^b$ where $a \gt b$,and $x$ any positive quantity,I don't know if this has a special name for it but we could prove this either by expanding or by just using Bernoulli's inequality. – Quixotic Sep 16 '11 at 20:10

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.

-

No AM-GM inequality - just simple computation:

\begin{align} \frac{(1+\frac{x}{n+1})^{n+1}}{(1+\frac{x}{n})^n} &= (1+\frac{x}{n})\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1} \\\\ &= (1+\frac{x}{n})\left(\frac{n(n+1)+nx}{(n+1)(n+x)}\right)^{n+1} \\\\ &= (1+\frac{x}{n})\left(\frac{(n+1)(n+x)-x}{(n+1)(n+x)}\right)^{n+1} \\\\ &= (1+\frac{x}{n})\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1} \\\\ &> (1+\frac{x}{n})(1-\frac{x}{n+x}) = \frac{n+x}{n} \frac{n}{n+x} = 1. \end{align}

Copied from a previous answer of mine.

-
Maybe you should note where you use bernoullis inequality. I personally like Aryabhata application of it more though. – Listing Dec 10 '11 at 9:35

This is equivalent to showing that $$\left(\frac{n}{n-1}\right)^{n-1}\tag{1}$$ is an increasing function of $n$. Consider the Taylor expansion of \begin{align} (n-1)\log\left(\frac{1}{1-1/n}\right) &=(n-1)\left(\frac1n+\frac12\frac1{n^2}+\frac13\frac1{n^3}+\dots\right)\\ &=1-\frac1{1\cdot2}\frac1n-\frac1{2\cdot3}\frac1{n^2}-\frac1{3\cdot4}\frac1{n^3}+\dots\tag{2} \end{align} $(2)$ is obviously an increasing function of $n$. QED

Another useful case

$$\left(\frac{n}{n-1}\right)^n\tag{3}$$ is a decreasing function of $n$. \begin{align} n\log\left(\frac{1}{1-1/n}\right) &=n\left(\frac1n+\frac12\frac1{n^2}+\frac13\frac1{n^3}+\dots\right)\\ &=1+\frac12\frac1n+\frac13\frac1{n^2}+\dots\tag{4} \end{align} $(4)$ is obviously a decreasing function of $n$.

A boundary case

$$\left(\frac{n}{n-1}\right)^{n-1/2}\tag{5}$$ is a decreasing function of $n$. \begin{align} (n-1/2)\log\left(\frac{1}{1-1/n}\right) &=(n-1/2)\left(\frac1n+\frac12\frac1{n^2}+\frac13\frac1{n^3}+\dots\right)\\ &=1+\frac12\left(\frac1{2\cdot3}\frac1{n^2}+\frac2{3\cdot4}\frac1{n^3}+\frac3{4\cdot5}\frac1{n^4}+\dots\right)\tag{6} \end{align} $(6)$ is obviously a decreasing function of $n$.

Comparing $(6)$ to $(2)$ shows that $\left(1+\frac1n\right)^{n+1/2}$ is a lot closer to $e$ than is $\left(1+\frac1n\right)^n$.

-