Proof verification of $\left(1+{1\over n}\right)^n > e^{1-1/n}$ for all $n\in\Bbb N$.

Question

Prove that for all $n\in\Bbb N$: $$ \left(1+{1\over n}\right)^n > e^{1-{1\over n}} $$

Suppose the inequality holds.

Using the fact that: $$ \lim_{n\to\infty}\left(1+{x\over n}\right)^n = e^x $$ and: $$ \left(1+{x\over n}\right)^n \le \left(1+{x\over n+1}\right)^{n+1} $$

put $x = {1-{1\over n}}$ in the above: $$ e^{1-{1\over n}} > \left(1+\frac{1-{1\over n}}{n}\right)^n = \\ = \left(1+\frac{n-1}{n^2}\right)^n = y_n $$

Let $x_n$: $$ x_n = \left(1+{1\over n}\right)^n $$

Consider the fraction: $$ \frac{x_n}{y_n} = \frac{\left(1+{1\over n}\right)^n}{\left(1+\frac{n-1}{n^2}\right)^n} \\ =\left(\frac{n^2(n+1)}{n(n^2 + n - 1)}\right)^n \\ = \left(\frac{n^2 + n}{n^2 + n - 1}\right)^n > 1 $$

Therefore $x_n > y_n$. But at the same time: $$ e^{1-{1\over n}} > y_n $$ Thus: $$ x_n > e^{1-{1\over n}} > y_n $$

Define: $$ z_n = e^{1-{1\over n}} $$

Now if we take the limit of both sides then by squeeze theorem we get: $$ \lim_{n\to\infty}x_n \ge\lim_{n\to\infty} z_n \ge \lim_{n\to\infty}y_n $$

All three limits are equal to $e$ which shows that $z_n$ is squeezed between $x_n$ and $y_n$ meaning that $x_n > z_n$.

I'm not pretty sure the reasoning above is valid, so i'm kindly asking to verify it or point to the flaws. In case the above is wrong could you please show the way to prove the inequality? Thank you!

Answer 1

Your flaw is that you are assuming the result to begin your proof. Here is a simple approach: $$\dfrac 1n +n\ln\left(1+\frac 1n\right)>\frac 1n+\frac{n}{n+1} = \dfrac{n^2+n+1}{n^2+n}>1.$$ Now, check that above is equivalent to your inequality. We used the well-known fact that: $$\ln(1+x)>\dfrac{x}{x+1}$$ for $x>0.$

Answer 2

Why should the squeeze theorem imply that $x_n>z_n$? The theorem says that

if $x_n>z_n>y_n$ and $\lim_{n\to\infty}x_n=l=\lim_{n\to\infty}y_n$, then also $\lim_{n\to\infty}z_n=l$.

However, it does not “reverse”. To see why, swap $z_n$ and $y_n$ in the statement above.

The inequality you have is equivalent to $$ n\log\Bigl(1+\frac{1}{n}\Bigr)>1-\frac{1}{n} $$ or, setting $x=1/n$, to $$ \log(1+x)>x-x^2 $$ for $0<x\le 1$. Consider $f(x)=\log(1+x)-x+x^2$ defined over $(-1,1]$; then $f(0)=0$ and $$ f'(x)=\frac{2x^2+x}{1+x}>0 $$ proving that $f(x)>0$ for $0<x\le1$.

Answer 3

Here is a sketch of another approach: $\frac{1}{1+1/n}\le \frac{1}{t}\le 1$ as soon as $t\in [1,1+1/n].$ Integrate across this inequality from $1$ to $1+1/n$ and find that $\frac{1}{n+1}\le \ln(1+1/n)\le 1/n.$ Now exponentiate and raise to the $n+1$ power, and simplify, and find that $\left ( 1+1/n \right )^{n}\le e\le \left ( 1+1/n \right )^{n+1}.$ The result follows easily from this.

Answer 4

Applying $\ln$ to both sides, we see this holds iff

$$n\ln(1+1/n) > 1-1/n$$

for $n\in \mathbb N.$ This is the same as saying

$$\frac{\ln(1+1/n)-\ln 1}{1/n}> 1-1/n.$$

By the MVT, the left side equals $1/c_n,$ where $c_n\in (1,1+1/n).$ Since $1/c_n> 1/(1+1/n)$ and $1>(1+1/n)(1-1/n)$ for all $n\in \mathbb N,$ we're done.