Lipossig

2020-12-28

Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series.

$\sum _{n=1}^{\mathrm{\infty}}\frac{1}{\sqrt{{n}^{3}+2n}}$

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${b}_{n}=\frac{1}{{n}^{\frac{3}{2}}}$

$\underset{n\to \mathrm{\infty}}{lim}\frac{{a}_{n}}{{b}_{n}}=\underset{n\to \mathrm{\infty}}{lim}\frac{\frac{1}{\sqrt{{n}^{3}+2n}}}{\frac{1}{{n}^{\frac{3}{2}}}}$

$=\underset{n\to \mathrm{\infty}}{lim}(\frac{1}{\sqrt{{n}^{3}+2n}}\times {n}^{\frac{3}{2}})$

$=\underset{n\to \mathrm{\infty}}{lim}(\frac{1}{{n}^{\frac{3}{2}}\sqrt{1+\frac{2}{{n}^{2}}}}\times {n}^{\frac{3}{2}})$

$=\underset{n\to \mathrm{\infty}}{lim}\left(\frac{1}{\sqrt{1+\frac{2}{{n}^{2}}}}\right)$

$=1$

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.

Suppose that we have two series

Where L is positive and finite then either both series are convergent or both series divergent.

Given that,

Since we have,

Here

So,

Now evaluate

Since

Since using p-series test [

Since

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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