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}}

Series

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

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