Question

Determine the convergence or divergence of the series. sum_{n=2}^inftyfrac{1}{n(ln n)^3}

Series
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asked 2020-11-06
Determine the convergence or divergence of the series.
\(\sum_{n=2}^\infty\frac{1}{n(\ln n)^3}\)

Answers (1)

2020-11-07
Consider the series
\(\sum_{n=2}^\infty\frac{1}{n(\ln n)^3}\)
Cauchy Condensation test:
Let \(\) be a decreasing sequence of positive real numbers. Then the two series \(\sum_{n=1}^\infty c_n\) and \(\sum_{n=1}^\infty2^nc_{2^n}\) either convergent or divergent.
Here
\(\sum_{n=2}^\infty\frac{1}{n(\ln n)^3}\)
\(=\frac{1}{n(\ln n)^3}\) monotonically decreasing series of positive real numbers for \(n\geq2\)
\(\sum_{n=2}^\infty c_n=\sum_{n=2}^\infty\frac{1}{n(\ln n)^3}\)
\(\sum_{n=2}^\infty 2^nc_{2^n}=\sum_{n=2}^\infty2^n\frac{1}{2^n(\ln2^n)^3}=\sum_{n=2}^\infty\frac{1}{(n\ln2)^3}=\sum_{n=2}^\infty\frac{1}{(\ln2)^3n^3}\)
\(\sum_{n=2}^\infty\frac{1}{n^3}\) is convergent by p-series test
\(\Rightarrow\sum_{n=1}^\infty2^nc_{2^n}=\frac{1}{(\ln2)^3}\sum_{n=2}^\infty\frac{1}{n^3}\) is convergent
Hence by Cauchy Condensation test \(\sum_{n=1}^\infty c_n\) and \(\sum_{n=1}^\infty 2^nc_{2^n}\) converge or diverge together.
Here \(\sum_{n=1}^\infty 2^nc_{2^n}=\frac{1}{(\ln2)^3}\sum_{n=2}^\infty\frac{1}{n^3}\) convergent implies \(\sum_{n=2}^\infty c_n=\sum_{n=2}^\infty\frac{1}{n(\ln n)^3}\) is convergent.
Answer: Convergent
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