36
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Why not try taking derivative for $f(x) = \displaystyle (1 + \frac{1}{x})^x$? $\endgroup$ – Shuhao Cao Sep 15 '11 at 19:25
  • 2
    $\begingroup$ See also: math.stackexchange.com/questions/51906/… $\endgroup$ – Aryabhata Sep 15 '11 at 20:50
  • 2
    $\begingroup$ @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? :)) $\endgroup$ – Srivatsan Sep 15 '11 at 20:54
  • $\begingroup$ @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. $\endgroup$ – Aryabhata Sep 15 '11 at 21:20
  • 4
    $\begingroup$ @Srivatsan: Since there are multiple answers without AM/GM and your comment has upvotes, I have added an answer. $\endgroup$ – Aryabhata Sep 16 '11 at 2:02
47
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – marty cohen Sep 15 '11 at 23:04
  • $\begingroup$ @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. $\endgroup$ – Srivatsan Sep 15 '11 at 23:07
23
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ +1, and it's fun to see something other than the AM-GM inequality sometimes. :) $\endgroup$ – Mike Spivey Sep 15 '11 at 20:06
  • $\begingroup$ @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. $\endgroup$ – user940 Sep 15 '11 at 21:19
  • $\begingroup$ No worries. I thought your argument was cleaner, so I deleted my answer. $\endgroup$ – Mike Spivey Sep 15 '11 at 22:20
22
$\begingroup$

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...).

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Quixotic Sep 16 '11 at 20:10
  • $\begingroup$ @Aryabhata: I don't follow on the part where you eliminate the quality. Can you elaborate how you got to that last equation? And expand on the "relatively prime" part. $\endgroup$ – Parseval Oct 23 '17 at 20:45
  • $\begingroup$ @Parseval: $(1 + \frac{1}{n+1}) = \frac{n+2}{n+1}$. Now transfer the denominators to the other side. Similarly with the denominator on the right. As to relatively prime, if a prime divides the left side, then it must divide the right side, and so must divide $n+1$ which would imply they are not relatively prime. $\endgroup$ – Aryabhata Oct 24 '17 at 2:12
13
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Maybe you should note where you use bernoullis inequality. I personally like Aryabhata application of it more though. $\endgroup$ – Listing Dec 10 '11 at 9:35
4
$\begingroup$

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 $\left(1+\frac1n\right)^n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.