Question

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

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

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.