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}}

Question
Series
asked 2021-01-16

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

Answers (1)

2021-01-17
Apply integral test and check weather series is convergeny or divergent.
If the term series \(a_n\) can be represented by a positive, decreasing, continuous functionm \(f(n)\) then,
If \(\lim_{R\to\infty}\int_a^R f(x)dx\) exists then \(\sum_{n=1}^\infty a_n\) converges.
If \(\lim_{R\to\infty}\int_a^R f(x)dx\) does not exist, then \(\sum_{n=1}^\infty a_n\) diverges.
Consider the given series:
\(S_n=\sum_{n=2}^\infty\frac{1}{n\sqrt{\ln n}}\)
Determine the convergence of the following Integral:
\(\int_2^\infty\frac{1}{x\sqrt{\ln x}}dx=\lim_{t\to\infty}\int_{2}^t\frac{1}{x\sqrt{\ln x}}dx\)
Apply u-substitution: \(u=\ln(x)\)
\(\int_2^t\frac{1}{x\sqrt{\ln x}}dx=\int_{\ln(2)}^{\ln(t)}\frac{1}{\sqrt{u}}du\)
\(=\int_{\ln(2)}^{\ln(t)}u^{-\frac{1}{2}}du\)
\(=[\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}]_{\ln(2)}^{\ln(t)}\)
\(=[2\sqrt{u}]_{\ln(2)}^{\ln(t)}\)
\(=2\sqrt{\ln(t)}-2\sqrt{\ln(2)}\)
Find whether series diverges.
Applying Limits:
\(\lim_{t\to\infty}\int_2^t\frac{1}{x\sqrt{\ln x}dx}=\lim_{t\to\infty}[2\sqrt{\ln(t)}-2\sqrt{\ln(2)}]\)
\(=[\infty-2\sqrt{\ln(2)}]\)
\(=\infty\)
=Diverges
Hence, \(\sum_{n=2}^\infty\frac{1}{n\sqrt{\ln n}}\) diverges
0

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