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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?

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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

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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.

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

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

$=(1/2)$;

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

$\sum a_n$ does not convergent.

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