let $a_k=\dfrac{1}{k\ln k}$

I found out that $a_{n+1} < a_n$ for all $n$.

Also, $\displaystyle\lim_{n\to \infty} a_n = 0$.

So by leibniz alternating series test, I said $\displaystyle\sum_{k=2}^{\infty}\frac{(-1)^{k+1}}{k\ln k}$ converges.

Is that true? Does it Absolute converge?


As you noted, it converges by Leibniz test. However, it does not converge absolutely. (Try the integral test on $\sum |a_k|$.)


It does converge because it is alternating and elements decrease monotonically to 0. It does not converges absoluteley, see the condensation test.

I hope this helps ;-)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.