# For any positive integer $n$, prove the following inequality. [duplicate]

Prove that for any positive integer n, the following inequality is true. $$\left(1+ \frac{1}{n}\right)^n < \left(1+\frac{1}{n+1}\right)^{n+1}$$

Attempt

Not a good attempt but this is my thinking

It will not be a problem when checking whether this is true for smaller integers. Only for larger integers, it will be difficult to prove. So, I took the equation LHS-RHS < $$0$$. I took limit for n tending to infinity and then applied L'Hospital Rule( though the equation was difficult to handle) & therefore could not do anything further.

Adding graph for @Rebellos 's last equation

## marked as duplicate by Jack D'Aurizio, Martin R, Don Thousand, Leucippus, user10354138Nov 6 '18 at 0:19

• Any hints and tips are welcome. – jayant98 Nov 5 '18 at 18:17
• Try the AM-GM inequality. This has been asked uncountably times, by the way. – Jack D'Aurizio Nov 5 '18 at 18:24
• – Martin R Nov 5 '18 at 20:24

If you are allowed to take derivatives, then letting $$f(x)=\left(1+\frac{1}{x}\right)^x$$, we have $$f'(x)=f(x)\left(\frac{(x+1)\log\left(1+\frac{1}{x}\right)-1}{x+1}\right)$$

And $$f'(x)>0$$ as long as $$x>0$$.

An alternative approach :

Assume it indeed is :

$$\bigg(1+ \frac{1}{n}\bigg)^n < \bigg(1+\frac{1}{n+1}\bigg)^{n+1}$$

Then, since the natural logarithm is a strictly increasing function, it is :

$$\ln\bigg(1+ \frac{1}{n}\bigg)^n < \ln\bigg(1+\frac{1}{n+1}\bigg)^{n+1} \Leftrightarrow n\ln\bigg(1+\frac{1}{n}\bigg) < (n+1)\ln\bigg(1+\frac{1}{n+1}\bigg)$$

$$n\bigg[\ln\bigg(1 + \frac{1}{n+1}\bigg)-\ln\bigg(1+\frac{1}{n}\bigg)\bigg]+\ln\bigg(1+\frac{1}{n+1}\bigg) > 0$$

$$\Leftrightarrow$$

$$\ln\Bigg(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\Bigg)^n + \ln\bigg(1+ \frac{1}{n+1}\bigg) >0$$

But for $$n \in \mathbb N^+$$ we can clearly see that this is true (why ?). Thus the initial inequality holds, since the logical $$\Leftrightarrow$$ implies it.

• @cansomeonehelpmeout Thanks for the typo correction ! – Rebellos Nov 5 '18 at 18:38
• Thanks for helping me but I have a doubt from 3rd step onwards. The 3rd term in the 3rd step shouldn't be positive and hence for the further steps also? – jayant98 Nov 5 '18 at 19:07
• @jayant98 Yes and this just made our life easier! – Rebellos Nov 5 '18 at 19:10
• @jayant98 Did you manage to prove why the last expression holds ? – Rebellos Nov 5 '18 at 19:24
• Nope,still trying. :| – jayant98 Nov 5 '18 at 19:26

I know this question is already answered, but I want to give an answer that only uses the binomium of Newton. This is certainly not the easiest way to show the inequality, but it is an "artisanal" way.

By the binomium we get \begin{align*} \left(1+ \frac{1}{n}\right)^n &= \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} \\ &= 1 + \frac{n}{1!}\frac{1}{n} + \frac{n(n-1)}{2!}\frac{1}{n^2} + \frac{n(n-1)(n-2)}{3!}\frac{1}{n^3}+ \ldots + \frac{n(n-1)(n-2)\cdots 1}{n!}\frac{1}{n^n} \\ &= 2 + \frac{1}{2!}\left(1-\frac{1}{n}\right)+ \frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+ \cdots + \frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{n-1}{n}\right). \end{align*} Similarly \begin{align*} \left(1 + \frac{1}{n+1}\right)^{n+1} &= 2 + \frac{1}{2!}\left(1-\frac{1}{n+1}\right)+ \frac{1}{3!}\left(1-\frac{1}{n+1}\right)\left(1-\frac{2}{n+1}\right)+ \cdots \\ & + \frac{1}{n!}\left(1-\frac{1}{n+1}\right)\left(1-\frac{2}{n+1}\right)\cdots\left(1-\frac{n-1}{n+1}\right) \\ &+ \frac{1}{(n+1)!}\left(1-\frac{1}{n+1}\right)\left(1-\frac{2}{n+1}\right)\cdots\left(1-\frac{n}{n+1}\right). \end{align*}

For every term in the expression for $$\left(1+\tfrac{1}{n}\right)^n$$ there is a similar term in the expression for $$\left(1+\tfrac{1}{n+1}\right)^{n+1}$$ that is equal of bigger. Furthermore, the last expression contains one positive term more. Hence the latter expression is bigger.

Since $$\log$$ function is strictly increasing, we have that

$$(1+ \frac{1}{n})^n < (1+\frac{1}{n+1})^{n+1} \iff n\log\left(1+\frac1n\right)<(n+1)\log\left(1+\frac1{n+1}\right)$$

and for $$x>1$$

$$f(x)=x\log x \implies f'(x)=\log x+1>0$$