Finding limit of the sequence

Given a sequence defined as

$${a_{n}}= \frac{1} {\sqrt{2}} + \frac{2} {3} + \frac{3} {\sqrt{28}} + \cdots \cdots + \frac{n} {\sqrt{n^3+1}}$$

and I want to find the limit of this sequence.

I think the limit is positive infinity, but I don't know what I should do in order to prove this.

Please give me some hints on how to approach this, thanks to anybody who helps.

Answer

$$\frac{n}{\sqrt{n^3+1}}\sim\frac1{\sqrt{n}},$$
hence

$$\lim_{n\to\infty}a_n=\infty.$$

Explanation, using series' methods

$$\lim_{n\to\infty}\frac{\dfrac{n}{\sqrt{n^3+1}}}{\dfrac1n}=1$$
and the series

$$\lim_{n\to\infty}\frac1{\sqrt{n}}$$
is divergent, hence also
$$\lim_{n\to\infty}\frac{n}{\sqrt{n^3+1}}$$
id divergent, but $a_n$'s are partial sums of this series.

More elementary explanation

It is easy to see, that
$$\dfrac{n}{\sqrt{n^3+1}}\geq\frac1n$$
for all $n\geq2$. Let
$$b_n=1+\frac12+\dots+\frac1n.$$
It is certainly increasing, hence it remains to show, that it is unbounded. Let us consider

$$b_{2^n}=1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\dots+\frac18\right)+\dots +\left( \frac1{2^{n-1}+1}+\dots+\frac1{2^n}\right).$$
But the value in every parentheses is greater than $\frac12$, hence $$b^{2^n}>1+\frac{n}{2}.$$