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 $x\in\mathbb{R}_+$,


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 $\frac{1}{4}$?

  • 3
    $\begingroup$ You are both right since $1/8<1/4$. $\endgroup$
    – Martin R
    Aug 23 '21 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.


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