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How to check the convergence of the series $$ \sum_{n=1}^\infty \frac{\ln\left(1+\frac{1}{n}\right)}{n^a} $$ I tried using all the tests: D'Alembert's, ratio test, comparison test but I failed . how to find when the series converges? Thanks in advance.

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Recall that the logarithm function satisfies the inequalities

$$\frac{x}{1+x}\le \log(1+x) \le x \tag 1$$

for all $x>0$. Letting $x=1/n$ in $(1)$, we see that

$$\sum_{n=1}^N \frac{1}{n^\alpha (1+n)} \le \sum_{n=1}^N \frac{\log\left( 1+\frac1n\right)}{n^\alpha}\le \sum_{n=1}^N \frac{1}{n^{1+\alpha}}\tag 2$$

Hence, the right-side inequality in $(2)$ shows that the series of interest converges for $\alpha>0$ and the left-side inequality shows that it diverges for $\alpha \le0$.

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  • $\begingroup$ Thanks for your attention. $\endgroup$ – math is fun May 24 '17 at 3:48
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    $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola May 24 '17 at 14:05
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(Detailed) Hint: When $n\to\infty$, we have $\frac{1}{n}\xrightarrow[n\to\infty]{}0$, and from there $\ln\left(1+\frac{1}{n}\right)\sim_{n\to\infty}{} \frac{1}{n}$. (In the sense of asymptotic equivalents, that is here that $$\frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}\xrightarrow[n\to\infty]{}1 $$ in other terms, $\ln\left(1+\frac{1}{n}\right)$ behaves asymptotically exactly like $\frac{1}{n}$).

By the (limit) comparison test, it follows that $\sum_{n=1}^\infty \frac{\ln\left(1+\frac{1}{n}\right)}{n^a}$ and $\sum_{n=1}^\infty \frac{\frac{1}{n}}{n^a}=\sum_{n=1}^\infty \frac{1}{n^{a+1}}$ have same nature (the first converges iff the second does). Now, the second is a $p$-series, so you should be able to handle it quite easily...

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Prologue. $1/(n+1)<\log (1+1/n)<1/n$ because $$1/(n+1)=\int_{1-1/(n+1)}^11dt<\int_{1-1/(n+1)}^1(1/t)dt=\left|\log (1-1/(n+1)\right|=$$ $$=\log (1+1/n)=$$ $$=\int_1^{1+1/n}(1/t)dt<\int_1^{1+1/n}1dt=1/n.$$ If $a>0$ then $0< n^{-a}\log (1+1/n)<n^{-a}(1/n)=1/n^{a+1}$ and we have convergence.

If $a\leq 0$ then $n^{-a}\geq 1$ so $n^{-a}\log(1+1/n)\geq \log(1+1/n)>1/(n+1)$ and we have divergence.

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  • $\begingroup$ Thanks for your attention. $\endgroup$ – math is fun May 24 '17 at 3:48

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