For any positive integer $n$, prove the following inequality.

Question

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.

Edit: I, now, want to know about @Rebellos' last equation and his answer (though at first my intention was to know about the question). Please see the comments of @Rebellos ' answer.

Adding graph for @Rebellos 's last equation

Answer 1

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 $x>0$.

Answer 2

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.

Answer 3

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 $x>1$

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

Answer 4

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.