# Series convergency

I have to prove that this series is convergent:

$$\sum_{i=1}^\infty \frac{\sqrt {n^2+1} -1 }{\sqrt[3]n}$$

I try to estimate, that

$$\ \frac{\sqrt {n^2+1} -1 }{\sqrt[3]n}~~is ~similar~to ~ \frac{1}{n^2}$$

I got

$$\ \frac{1}{n\sqrt[3] {\frac {1}{n^2}}n \sqrt{1+ \frac{1}{n^2}} - n*n \sqrt {\frac{1}{n^2}}}$$

Firstly, I am not convinced that I assumed properly the similarity in this formula.

Secondly, I suppose that I have to compare the series with another series bigger than this and convergent. Is it a good idea to use:

$$\ \frac{\sqrt {n^2+2} -1 }{\sqrt[3]{n+1}} ?$$

EDIT: Thank you for your all contributions; however, I have realised my mistake. It should have been:

$$\sum_{i=1}^\infty \frac{\sqrt {n^2+1} -n }{\sqrt[3]n}$$

My previous calculations apply to this example. Could you help to solve it?

The series is divergent: since $$\sqrt {n^2 + 1} - 1 \sim n\,\,\,\,\,\,\,\left( {n \to + \infty } \right)$$ it is $$a_n = \frac{{\sqrt {n^2 + 1} - 1}} {{\sqrt[3]{n}}} \sim n^{2/3} \,\,\,\,\,\left( {n \to + \infty } \right)$$ so the necessary condition for the convergence $$\mathop {\lim }\limits_{n \to + \infty } a_n = 0$$ is not satisfied and the series is divergent. Therefore

Since\begin{align}\require{cancel}\lim_{n\to\infty}\frac{\sqrt{n^2+1}-1}{\sqrt[3]n}&=\lim_{n\to\infty}\frac{n^2}{\sqrt[3]n\left(\sqrt{n^2+1}+1\right)}\\&=\lim_{n\to\infty}n^{2/3}\frac{\cancel{n^{1/3}}n}{\cancel{\sqrt[3]n}\left(\sqrt{n^2+1}+1\right)}\\&=\lim_{n\to\infty}n^{2/3}\frac1{\sqrt{1+\frac1{n^2}}+\frac1n}\\&=\infty,\end{align}the series diverges.

$$a_n:=\dfrac {\sqrt{n^2+1}-1}{\sqrt[3]{n}}\ge \dfrac{\sqrt{n+0}-1}{\sqrt[3]{n}}$$

$$= \dfrac{√n-1}{\sqrt[3]{n}}\ge \dfrac{√n-(1/2)√n}{√n}$$

$$=(1/2)$$;

$$\lim_{ n \rightarrow \infty}a_n\not =0$$.

$$\sum a_n$$ does not convergent.