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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $$ \sum_{n\geq 1}\frac{\log(n+1)}{n(n+1)}=\int_{0}^{+\infty}\frac{(e^{-x}-1)\log(1-e^{-x})}{x}\,dx=\int_{0}^{1}\frac{(1-x)\log(1-x)}{x\log x}\,dx. $$

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

implies

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

is simple and based on the comparison test.

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