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?

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    $\begingroup$ Why not try taking derivative for $f(x) = \displaystyle (1 + \frac{1}{x})^x$? $\endgroup$
    – Shuhao Cao
    Sep 15 '11 at 19:25
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    $\begingroup$ See also: math.stackexchange.com/questions/51906/… $\endgroup$
    – Aryabhata
    Sep 15 '11 at 20:50
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    $\begingroup$ @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
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    $\begingroup$ @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

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

  • $\begingroup$ 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. $\endgroup$ 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

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.

  • $\begingroup$ +1, and it's fun to see something other than the AM-GM inequality sometimes. :) $\endgroup$ 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. $\endgroup$ 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...).

  • $\begingroup$ 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

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.

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    $\begingroup$ 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

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


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