Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

Cabiolab
2020-10-18
Answered

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

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tabuordy

Answered 2020-10-19
Author has **91** answers

Integral test:

Let f(x) is positive decreasing continuous function on$[k,\mathrm{\infty})$ and $f\left(n\right)={a}_{n}$ then

If${\int}_{k}^{\mathrm{\infty}}f\left(x\right)dx$ is convergent then $\sum _{n=k}^{\mathrm{\infty}}{a}_{n}$ is also convergent.

If${\int}_{k}^{\mathrm{\infty}}f\left(x\right)dx$ is divergent then $\sum _{n=k}^{\mathrm{\infty}}{a}_{n}$ is also divergent.

So we can use integral test in given function.

Given that

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

Integral test:

Here$f\left(x\right)=\frac{1}{{(2n+3)}^{3}}$ so

$\int}_{1}^{\mathrm{\infty}}\frac{1}{{(2n+3)}^{3}}dx={\int}_{1}^{\mathrm{\infty}}{(2x+3)}^{-3}dx=\frac{1}{2}{\left[\frac{{(2x+3)}^{-2}}{-2}\right]}_{1}^{\mathrm{\infty}$

$=-\frac{1}{4}{\left[\frac{1}{{(2x+3)}^{2}}\right]}_{1}^{\mathrm{\infty}}$

$=-\frac{1}{4}[\frac{1}{25}-0]$

$=\frac{1}{100}$

Which is finite, so${\int}_{1}^{\mathrm{\infty}}\frac{1}{{(2x+3)}^{3}}dx$ is convergent

Since${\int}_{1}^{\mathrm{\infty}}\frac{1}{{(2x+3)}^{3}}dx$ is convergent so using integral test $\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{(2n+3)}^{3}}$ si also convergent.

Let f(x) is positive decreasing continuous function on

If

If

So we can use integral test in given function.

Given that

Integral test:

Here

Which is finite, so

Since

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