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# 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.sum_{n=1}^inftyfrac{1}{n+3}

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asked 2021-03-08

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.
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{n}+{3}}}}$$

## Answers (1)

2021-03-09
Integral Test:
If f(x) is continuous, positive and decreasing function on $$\displaystyle{\left[{k},\infty\right]}$$ and $$\displaystyle{f{{\left({n}\right)}}}={a}_{{n}}$$ then
If $$\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$ is convergent then $$\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$$ is also convergent.
If $$\displaystyle{\int_{{k}}^{\infty}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$ is divergent then $$\displaystyle{\sum_{{{n}={k}}}^{\infty}}{a}_{{n}}$$ is also divergent.
Given that
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{n}+{3}}}}$$
Using Integral test,
Here $$\displaystyle{f{{\left({x}\right)}}}={a}_{{n}}={\frac{{{1}}}{{{x}+{3}}}}$$
$$\displaystyle{\int_{{1}}^{\infty}}{\frac{{{1}}}{{{x}+{3}}}}{\left.{d}{x}\right.}={{\left[{\ln{{\left({n}+{3}\right)}}}\right]}_{{1}}^{\infty}}$$
$$\displaystyle={\ln{{\left(\infty\right)}}}+{\ln{{4}}}$$
$$\displaystyle=\infty$$
Since integral $$\displaystyle{\int_{{1}}^{\infty}}{\frac{{{1}}}{{{x}+{3}}}}{\left.{d}{x}\right.}$$ is divergent so using integral test $$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{n}+{3}}}}$$ is also divergent.

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