Consider the series $$\sum_{n=1}^\infty\frac{(-1)^n}{n\log^2(n+1)}.$$ Determine whether it converges absolutely or conditionally.

My attempt

S=$\sum_{n=1}^{\infty}( -1)^n$ an

an is monotonically decreasing and it approaches zero when n approaches infinity. So series is convergent .


How to check for absolute convergence? Ratio test fails here.

  • 1
    $\begingroup$ Is the series $\sum_{n=1}^\infty\frac{(-1)^n}{n\log^2(n+1)}$? $\endgroup$ – José Carlos Santos Jul 30 '18 at 10:12
  • $\begingroup$ To test the absolute convergence, you could try Cauchy's integral test. $\endgroup$ – xbh Jul 30 '18 at 10:15
  • $\begingroup$ Heh. When I first read this question, I assumed that $\log^2$ was the iterated logarithm ($\log\circ\log$), not the square. I am not sure if there is any firmly established convention in this area. $\endgroup$ – Harald Hanche-Olsen Jul 30 '18 at 11:18

For the absolute convergence by cauchy condensation test we can consider the convergenge of the condensed series $\sum 2^n a_{2^n}$ that is

$$\sum \frac{2^n}{2^n(\log^2(2^n+1))}=\sum \frac{1}{\log^2(2^n+1)}$$

which converges by limit comparison test with $\sum \frac1{n^2}$ indeed

$$\frac{1}{\log^2(2^n+1)}\sim\frac1{n^2\log^2 2}$$


Yes, you are correct, ratio test fails here. Hint: note that for $n\geq 3$, $$0\leq \frac{1}{n(\log^2(n+1))}\leq \frac{1}{n(\log^2(n))}\leq \int_{n-1}^{n}\frac{dx}{x(\log^2(x))}=\frac{1}{\log(n-1)}-\frac{1}{\log(n)}.$$

What may we conclude?


Simply we have $$\sum_{n=2}^\infty\frac{1}{n\log^2(n+1)}<\sum_{n=2}^\infty\frac{1}{n\log^2(n)}$$ and one may use the integral test for evaluating $\displaystyle\int_{2}^\infty\frac{dx}{x\log^2x}=\dfrac12$.


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.