I need some help here. I can show that if the Cauchy condensation test holds, then I get two separate series, one which converges by the comparison test, and one that converges by the ratio test. But I don't even know if this is a valid argument, since I'm not sure how to even check that the terms are decreasing. So I don't think this approach works.

I see that the same question has been asked here, but I'm not really satisfied with the answers. Are there simple ways to determine convergence, with something like the comparison test?

  • 2
    $\begingroup$ Both Mark Viola's answer and DeepSea's answer in the link provides a solution using comparison test. Also, limit comparison test with $\frac{1}{n^2}$ is valid. $\endgroup$ – Sangchul Lee Oct 12 '18 at 4:05
  • $\begingroup$ @Lee But what if I I'm in a test taking scenario and don't know how, or don't have time, to prove that $\log (x+1) < x$ for $x>1$? $\endgroup$ – Wesley Oct 12 '18 at 4:09
  • 1
    $\begingroup$ Then I recommend using limit comparison test together with the knowledge that $\lim_{x\to0} \frac{\log(1+x)}{x} = 1$. Here, the statement of limit comparison test is as follows: Let $(a_n)$ and $(b_n)$ be sequences of positive real numbers such that $a_n/b_n$ converges to a number in $(0, \infty)$. Then $\sum a_n$ converges if and only if $\sum b_n$ converges. $\endgroup$ – Sangchul Lee Oct 12 '18 at 4:11
  • $\begingroup$ $\displaystyle{\ln\left(1 + 1/n\right) \over n} \sim {1 \over n^{2}}$ as $\displaystyle n \to \infty$. So ?. $\endgroup$ – Felix Marin Oct 12 '18 at 20:47

By summation by parts

$$ \sum_{n=1}^{N}\frac{\log(1+1/n)}{n}=\frac{\log(N+1)}{N}+\sum_{n=1}^{N-1}\frac{\log(n+1)}{n(n+1)} $$ and by the Cauchy-Schwarz inequality $\log(n+1)\leq \sqrt{n+1}-\frac{1}{\sqrt{n+1}}$, such that the rearranged/decelerated series $\sum_{n\geq 1}\frac{\log(n+1)}{n(n+1)}$ is blatantly absolutely convergent.

By Frullani's theorem we also have the integral representation $$ \sum_{n\geq 1}\frac{\log(n+1)}{n(n+1)}=\int_{0}^{+\infty}\frac{(e^{-x}-1)\log(1-e^{-x})}{x}\,dx=\int_{0}^{1}\frac{(1-x)\log(1-x)}{x\log x}\,dx. $$


$$\log(1+x)\le x$$


$$\sum_{n=1}^\infty\frac{\log\left(1+\dfrac1n\right)}{n}\le \sum_{n=1}^\infty\frac1{n^2}$$

is simple and based on the comparison test.


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.