Convergence or divergence of series where the terms include reciprocal of the natural logarithm

Question

Show whether the following series - $ \sum_1^{\infty}\frac{1}{n\ln(n)} $ converges or diverges. Is it possible to make in particular a clever use of the limit comparison test (described in the book on Calculus by Thomas and Finney)?

Answer 1

Because we know that $lim s_n$ exists, we can find the sum by calculating the limit of subsequence $s_{2^n}$ (also called Cauchy condensation test) $s_{2^n} = \sum\limits_{k=2}^{2^n} {1\over k ln(k)} = \sum\limits_{k=1}^{n} \sum\limits_{i=2^{k-1}+1}^{2^k} {1\over i ln(i)} \geq \sum\limits_{k=1}^{n} \sum\limits_{i=2^{k-1}+1}^{2^k} {1\over 2^{k} ln(2^{k})} = \sum\limits_{k=1}^{n} {{2^{k-1}}\over{2^k ln(2^k)}} = {1\over {2 ln(2)}}\sum\limits_{k=1}^{n} {1 \over k} \to +\infty$.

Therefore, the series diverges.

EDIT: If you're not comfortable with summations:

$s_{2{^n}} = ({1\over{2ln(2)}}) + ({1\over{3ln(3)}} + {1\over{4ln(4)}}) +({1\over{5ln(5)}} + ... + {1\over{8ln(8)}}) + ... \geq {1\over{2 ln(2)}} + {2\over{4 ln(4)}} + {4\over{8 ln(8)}} + ... = {1\over {2ln(2)}}(1 + {1\over2} + {1\over3} + ...) \to \infty$

Answer 2

The sequence $\left(\frac1{n\ln n}\right)_{n\ge3}$ is non negative decreasing so by the integral test the series $$\sum_{n\ge2}\frac1{n\ln n}$$ is divergent since the integral $$\int_2^\infty\frac{dx}{x\ln x}=\ln(\ln(x)\bigg|_2^\infty=\infty$$