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}

Question
Series
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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