# Why is $\frac{n}{2(n+1)^2}\leq\frac{1}{4}$?

I have the following exercise in my textbook and I'm not completely sure about one thing in the answer:

Denote $$\mathbb{R}_{+}=(0,\infty)$$. Consider the probability space $$\mathbb{R}_+,\mathcal{B}(\mathbb{R}_+),P)$$ where $$P$$ is the exponentiat distribution $$dP(x)=e^{-x}\mathrm{dx}$$

Consider the random variables $$f_n(x)\exp\left(\frac{n}{2(n+1)^2)}x^{1/n}\right)$$

Where $$n\in\mathbb{N}$$.

Using the dominated convergence theorem, prove that the limit $$\lim_{n\to\infty}E(f_n)$$

exists and find it.

The answer to the question is the following:

We have $$\lim+{n\to\infty}\frac{n}{2(n+1)^2}=0$$

and for each $$x\in\mathbb{R}_+$$,

$$\lim_{n\to\infty}x^{1/n}=x^0=1$$

Therefore, for each $$x\in\mathbb{R}_+$$, $$\lim_{n\to\infty}f_n(x)=f(x),$$

Where $$f(x)=x.$$

For $$x\in\mathbb{R}_+$$, we have $$x^{1/n}\leq \text{max}\{1,x\}\leq 1+x$$

and $$\frac{n}{2(n+1)^2}\leq\frac{1}{4}$$

The answer continues further and I understand all the logic after it. However i'm not sure why is the last inequality true.

It seems to me that the maximum the LHS can attain is $$\frac{1}{8}$$, for $$n=1$$, since $$\frac{1}{2(4)}=\frac{1}{8}$$

Why would the author put the bound at $$\frac{1}{4}$$?

• You are both right since $1/8<1/4$. Aug 23 at 6:04

We have that

$$\frac{n}{2(n+1)^2}\le \frac{n+1}{2(n+1)^2}=\frac1{2(n+1)}\le \frac14$$

The author probably choose this bound because it suffices and it is simple to obtain.

Of course we can proceed observing that $$f(1)=\frac18$$ and then showing that $$f(n)$$ is decreasing.

$$(n-1)^{2} \geq 0$$ and this gives $$(n+1)^{2} \geq 4n$$. So $$\frac n {(n+1)^{2}} \leq \frac 1 4$$ and $$\frac n {2(n+1)^{2}} \leq \frac 1 8<\frac 1 4$$.

Since $$n^2 + 1 \ge 0$$ implies that $$n^2 + 2n + 1 \ge 2n$$.

So $$(n+1)^2 \ge 2n$$. Thus $$\frac12 \ge \frac{n}{(n+1)^2}$$ from which the inequality follows.