Lipossig

## Answered question

2020-12-28

Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{{n}^{3}+2n}}$

### Answer & Explanation

au4gsf

Skilled2020-12-29Added 95 answers

Limit comparison test:
Suppose that we have two series $\sum {a}_{n}$ and $\sum {b}_{n}$, define,
${b}_{n}=\frac{1}{{n}^{\frac{3}{2}}}$
Where L is positive and finite then either both series are convergent or both series divergent.
Given that,
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{{n}^{3}+2n}}$
Since we have,
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sqrt{{n}^{3}+2n}}$
Here
${a}_{n}=\frac{1}{\sqrt{{n}^{3}+2n}}$
So,
${b}_{n}=\frac{1}{{n}^{\frac{3}{2}}}$
Now evaluate ${b}_{n}=\frac{1}{{n}^{\frac{3}{2}}}$
$\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{\sqrt{{n}^{3}+2n}}}{\frac{1}{{n}^{\frac{3}{2}}}}$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{\sqrt{{n}^{3}+2n}}×{n}^{\frac{3}{2}}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{{n}^{\frac{3}{2}}\sqrt{1+\frac{2}{{n}^{2}}}}×{n}^{\frac{3}{2}}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{\sqrt{1+\frac{2}{{n}^{2}}}}\right)$
$=1$
Since $\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=L=1$ which is finite and positive.
Since using p-series test [$\sum \frac{1}{{n}^{p}}$ is convergent if $p>1$ or divergent if $p\le 1$], ${b}_{n}=\frac{1}{{n}^{\frac{3}{2}}}$ is convergent.
Since $\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}$ is finite and positive and $\sum {b}_{n}$ is convergent so using limit comparison test $\sum {a}_{n}$ is also convergent.

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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