How to show $\left(1-\frac{2}{n^2}\right)^{n^2/2} \le \frac1e$?

Question

How can you show that

$$\left(1-\frac{2}{n^2}\right)^{n^2/2} \le \frac{1}{e}\:\: \qquad\forall n \ge 2$$

Any ideas? Infinite series have never really been my thing. Thanks

Answer 1

Hints:

$$\left(1-\frac{2}{n^2}\right)^{n^2/2}=\left(1-\frac{1}{\frac{n^2}{2}}\right)^\frac{n^2}{2}$$

Now just use the fact that

$$\left(1-\frac{1}{f(n)}\right)^{f(n)}\xrightarrow[n\to\infty]{}e^{-1}$$

for any function $\,f\,$ s.t. $\,f(n)\xrightarrow[n\to\infty]{}\infty\,$ and, perhaps (depending on what you can assume), also the fact that your sequence is monotone ascending.

Answer 2

You know that $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n} = e$$ Hence, you know that $$\lim_{n\to\infty}\left[\left(1+\frac{1}{\frac{-n^2}{2}}\right)^{\frac{-n^2}{2}}\right]^{-1} = e^{-1} = \frac{1}{e}$$ You could calculate the derivate of this expression for example and show that it increases and then show that for $n=2$ the inequality holds. And that's it.

Answer 3

Just $\log$ both sides

$$ \log\left(1-\frac{2}{n^2}\right)^{n^2/2} \le \log \frac{1}{e}\:\: $$

$$ \frac 1 2 *n^2 *\log(1-\frac 2 {n^2}) \le -1\:\: $$

$$ \log(1-\frac 2 {n^2}) \le \frac {-2} {n^2}\:\: $$

$$-2/n^2 = t$$

$$ \log (1+t)\le t\qquad\forall t \ge -1/2$$

If last inequality not obivious for you just plot it and see that it works for $t > -1$