Question

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=2}^inftyfrac{1}{nsqrt{ln n}}

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asked 2021-02-08

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.
\(\displaystyle{\sum_{{{n}={2}}}^{\infty}}{\frac{{{1}}}{{{n}\sqrt{{{\ln{{n}}}}}}}}\)

Answers (1)

2021-02-09
Apply integral test and check weather series is convergeny or divergent.
If the term series \(\displaystyle{a}_{{n}}\) can be represented by a positive, decreasing, continuous functionm \(\displaystyle{f{{\left({n}\right)}}}\) then,
If \(\displaystyle\lim_{{{R}\to\infty}}{\int_{{a}}^{{R}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\) exists then \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}\) converges.
If \(\displaystyle\lim_{{{R}\to\infty}}{\int_{{a}}^{{R}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\) does not exist, then \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}\) diverges.
Consider the given series:
\(\displaystyle{S}_{{n}}={\sum_{{{n}={2}}}^{\infty}}{\frac{{{1}}}{{{n}\sqrt{{{\ln{{n}}}}}}}}\)
Determine the convergence of the following Integral:
\(\displaystyle{\int_{{2}}^{\infty}}{\frac{{{1}}}{{{x}\sqrt{{{\ln{{x}}}}}}}}{\left.{d}{x}\right.}=\lim_{{{t}\to\infty}}{\int_{{{2}}}^{{t}}}{\frac{{{1}}}{{{x}\sqrt{{{\ln{{x}}}}}}}}{\left.{d}{x}\right.}\)
Apply u-substitution: \(\displaystyle{u}={\ln{{\left({x}\right)}}}\)
\(\displaystyle{\int_{{2}}^{{t}}}{\frac{{{1}}}{{{x}\sqrt{{{\ln{{x}}}}}}}}{\left.{d}{x}\right.}={\int_{{{\ln{{\left({2}\right)}}}}}^{{{\ln{{\left({t}\right)}}}}}}{\frac{{{1}}}{{\sqrt{{{u}}}}}}{d}{u}\)
\(\displaystyle={\int_{{{\ln{{\left({2}\right)}}}}}^{{{\ln{{\left({t}\right)}}}}}}{u}^{{-{\frac{{{1}}}{{{2}}}}}}{d}{u}\)
\(\displaystyle={{\left[{\frac{{{u}^{{-{\frac{{{1}}}{{{2}}}}+{1}}}}}{{-{\frac{{{1}}}{{{2}}}}+{1}}}}\right]}_{{{\ln{{\left({2}\right)}}}}}^{{{\ln{{\left({t}\right)}}}}}}\)
\(\displaystyle={{\left[{2}\sqrt{{{u}}}\right]}_{{{\ln{{\left({2}\right)}}}}}^{{{\ln{{\left({t}\right)}}}}}}\)
\(\displaystyle={2}\sqrt{{{\ln{{\left({t}\right)}}}}}-{2}\sqrt{{{\ln{{\left({2}\right)}}}}}\)
Find whether series diverges.
Applying Limits:
\(\displaystyle\lim_{{{t}\to\infty}}{\int_{{2}}^{{t}}}{\frac{{{1}}}{{{x}\sqrt{{{\ln{{x}}}}}{\left.{d}{x}\right.}}}}=\lim_{{{t}\to\infty}}{\left[{2}\sqrt{{{\ln{{\left({t}\right)}}}}}-{2}\sqrt{{{\ln{{\left({2}\right)}}}}}\right]}\)
\(\displaystyle={\left[\infty-{2}\sqrt{{{\ln{{\left({2}\right)}}}}}\right]}\)
\(\displaystyle=\infty\)
=Diverges
Hence, \(\displaystyle{\sum_{{{n}={2}}}^{\infty}}{\frac{{{1}}}{{{n}\sqrt{{{\ln{{n}}}}}}}}\) diverges
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