Proving $\left[1+\frac{1}{(n+1)^2}\right]^{(n+1)^2}-\left[1+\frac{1}{n^2}\right]^{n^2}>0$ for $n\geq 2$, by induction

Question

I have to prove that $$\left[1+\frac{1}{(n+1)^2}\right]^{(n+1)^2}-\left[1+\frac{1}{n^2}\right]^{n^2}>0$$ for $n\geq 2$.

I checked the base case and it holds true for $n=2$.

Assume it is true for $n$ then,

$$\left[1+\frac{1}{(n+1)^2}\right]^{(n+1)^2}>\left[1+\frac{1}{n^2}\right]^{n^2}$$

How should I proceed from here. Any help is appreciated.

Answer 1

Observation:

Let's say

$a_n = (1 + \frac{1}{n})^n$

and

$b_n = (1 + \frac{1}{n^2})^{n^2}$

Obviously, $b_n = a_{n^2}$ , and we can easily prove that $a_n$ is increasing (see here).

So, $b_{n+1} = a_{(n+1)^2} > a_{n^2} = b_n$

Middle inequality holds because $(n+1)^2 > n$ and $a_n$ is increasing.

So, you are actually trying to prove that $a_n$ is increasing but in the set of natural squares, ie $1,4,9,16,25,\dots$

In order to use pure induction, the idea is to link your n-th term (which is basically $n^2$) with his previous natural number, ie $n^2-1$.

Doing so, we get

$(1+\frac{1}{n^2})^{n^2} > (1+\frac{1}{n^2-1})^{n^2-1} \Leftrightarrow$

$\Big(\frac{1+\frac{1}{n^2}}{1+\frac{1}{n^2-1}}\Big)^{n^2} > (1+\frac{1}{n^2-1})^{-1} \Leftrightarrow$

$(1-\frac{1}{n^4})^{n^2} > 1 - \frac{1}{n^2} $

Which holds because of the Bernoulli's inequality (see here)

Now, doing this step $2n-1$ times you get

$(1+\frac{1}{n^2})^{n^2} > (1+\frac{1}{n^2-2n+1})^{n^2-2n+1} \Leftrightarrow$

$(1+\frac{1}{n^2})^{n^2} > (1+\frac{1}{(n-1)^2})^{(n-1)^2}$

Answer 2

I'm not sure how one could proceed with induction, so I'll present an alternate proof that uses the methods of calculus. I learned it by solving Problem 90 from Section 11.1 of James Stewart's Calculus: Early Transcendentals (8e), and it definitely qualifies as one of my all-time favorite proofs!

Consider the sequence $a_n=\left(1+\frac{1}{n}\right)^n$. It's clear that $n^2<(n+1)^2$ for every $n\geq 1$, so if we can prove that $a_n$ is strictly increasing, that is

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

for all $n\geq 1$, your inequality immediately follows.

Given some integer $n\geq 1$, consider the function $f(x)=x^{n+1}$ defined over $[0,\infty)$. This function is differentiable over $(0,\infty)$, so given any pair of numbers $a,b\in\mathbb{R}$ with $0\leq a<b$, the mean value theorem implies that there is a $c\in(a,b)$ satisfying

$$\frac{f(b)-f(a)}{b-a}=f'(c)$$

that is,

$$\frac{b^{n+1}-a^{n+1}}{b-a}=(n+1)c^n$$

We know that $n\geq 1$, so $f'$ must be strictly increasing. Thus,

$$\frac{b^{n+1}-a^{n+1}}{b-a}=(n+1)c^n=f'(c)<f'(b)<(n+1)b^n$$

We now rearrange the resulting inequality

\begin{align*} \frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n &\iff b^{n+1}-a^{n+1}<(b-a)(n+1)b^n\\ &\iff b^{n+1}-(b-a)(n+1)b^n<a^{n+1}\\ &\iff b^{n}\left[b-(nb+b-na-a)\right]<a^{n+1}\\ &\iff b^{n}(b-nb-b+na+a)<a^{n+1}\\ &\iff b^{n}\left[(n+1)a-nb\right]<a^{n+1}\\ \end{align*}

Now we substitute $a=1+\frac{1}{n+1}$ and $b=1+\frac{1}{n}$ This is justified because the inequality holds for $0\leq a<b$ and $0<1+\frac{1}{n+1}<1+\frac{1}{n}$.

$$\left(1+\frac{1}{n}\right)^{n}\left[(n+1)\left(1+\frac{1}{n+1}\right)-n\left(1+\frac{1}{n}\right)\right]<\left(1+\frac{1}{n+1}\right)^{n+1}$$

$$\iff \left(1+\frac{1}{n}\right)^{n}\left[n+1+1-n-1\right]<\left(1+\frac{1}{n+1}\right)^{n+1}$$

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

This proves that $a_n$ is strictly increasing, which immediately implies that

$$\left(1+\frac{1}{n^2}\right)^{n^2}<\left(1+\frac{1}{(n+1)^2}\right)^{(n+1)^2}$$

or

$$\left(1+\frac{1}{(n+1)^2}\right)^{(n+1)^2}-\left(1+\frac{1}{n^2}\right)^{n^2}>0$$