Why is $\sum_{k=2}^{\infty} \frac{(-1)^k}{k\ln k}$ conditionally convergent?

Question

I was solving this question, and I thought it would be absolutely convergent because when you set the modulus, you would get $$\sum_{k=2}^{\infty} \frac{1}{k\ln k}.$$ But the answer says it is conditionally convergent, and I don't know why.

Answer 1

Because Cauchy-condensation criterion gives $$\sum_{k=2}^{\infty} \frac{1}{k\ln k} \sim \sum_{k=2}^{\infty} \frac{2^k}{2^k\ln 2^k} =\sum_{k=2}^{\infty}\frac{1}{k\ln 2}$$

Answer 2

Apply the integral test:\begin{align}\int_2^\infty\frac1{x\log x}\,\mathrm dx&=\int_2^\infty\frac{\log'x}{\log x}\,\mathrm dx\\&=\lim_{M\to\infty}\log\bigl(\log(M)\bigr)-\log\big(\log(2)\bigr)\\&=\infty.\end{align}

Answer 3

Or use Ermakoff's convergence test: $$\lim_\limits{x\to\infty}\frac{e^xf(e^x)}{f(x)}=\lim_\limits{x\to\infty}\frac{e^x\cdot \frac{1}{e^{x}\ln e^x}}{\frac{1}{x \ln x}}=\lim_\limits{x\to\infty}\ln x=\infty>1.$$ Also, see here.