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.
$$ \frac{n}{\sqrt{n^3+1}}\sim\frac1{\sqrt{n}}, $$ 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 $$ b^{2^n}>1+\frac{n}{2}. $$
Note that for $n > 1$
$$\frac{n}{\sqrt{n^3 +1}} = \frac{1}{\sqrt{n + 1/n^2 }}> \frac1{\sqrt{2n}}> \frac1{\sqrt{2}n},$$
and the series diverges because the harmonic series $\sum 1/n$ diverges.
