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I need to find if the following series converges absolutely: $$\sum_{n=2}^{\infty}(-1)^n\frac{\sqrt{n^2+1}-n}{\ln(n)}$$

I know that the series itself is a leibniz series and thus converges, but I don't know if it's converges absolutely.

The "new" series is $$\sum_{n=2}^{\infty}\bigg|(-1)^n\frac{\sqrt{n^2+1}-n}{\ln(n)}\bigg|=\sum_{n=2}^{\infty}\frac{\sqrt{n^2+1}-n}{\ln(n)}$$.

I tried to use the ratio test, but got $1$, and I can't find a series to compare with it.

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$$\frac{\sqrt{n^2+1}-n}{\ln(n)} = \frac{n \left(\sqrt{1 + \frac{1}{n^2}}-1\right)}{\ln(n)} = \frac{ \frac{1}{2n} + o \left( \frac{1}{n^2}\right)}{\ln(n)} \sim \frac{1}{2 n \ln (n)}$$

so your series is not absolutely convergent.

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$\sqrt{n^2+1}-n =\dfrac1{\sqrt{n^2+1}+n} \gt \dfrac1{2n+1} $.

Since $\sum \dfrac1{n\ln(n)} $ diverges, so does this.

To show that $\sum \dfrac1{n\ln(n)} $ diverges, run it through the Cauchy condenser, noting that $\sum_{n=2^{m}}^{2^{m+1}-1} \dfrac1{n\ln(n)} \ge \sum_{n=2^{m}}^{2^{m+1}-1} \dfrac1{2^m\ln(2^m)} =\dfrac{2^m}{2^mm\ln(2)} =\dfrac1{m\ln(2)} $ and the sum of these diverges.

Another proof of divergence is $(\ln\ln(x))' =\dfrac{(\ln(x))'}{\ln(x)} =\dfrac1{x\ln(x)} $ so $\ln\ln(x) =\int_e^x \dfrac{dt}{t\ln(t)} $.

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Since $\;\cfrac{\sqrt{n^2+1}-n}{\ln n}=\cfrac{1}{\left(\sqrt{n^2+1}+n\right)\ln n}>\cfrac{1}{3n\ln n}$

for all $\;n\in\mathbb{N}\;$ such that $\;n\ge2$

and $\;\sum\limits_{n=2}^\infty \cfrac{1}{3n\ln n}\;$ is divergent,

then, by applying comparison test, we obtain that

the series $\;\sum\limits_{n=2}^\infty\cfrac{\sqrt{n^2+1}-n}{\ln n}\;$ is divergent too.

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By Cauchy condensation test, for the condensed series we obtain

$$\frac{2^n\sqrt{2^{2n}+1}-2^{2n}}{\ln 2^n}=\frac{2^n\sqrt{2^{2n}+1}-2^{2n}}{n\ln 2}\,\frac{2^n\sqrt{2^{2n}+1}+2^{2n}}{2^n\sqrt{2^{2n}+1}+2^{2n}}=$$

$$=\frac{2^{2n}}{n\ln 2(2^n\sqrt{2^{2n}+1}+2^{2n})}=\frac{1}{n\ln 2\left(\sqrt{1+\frac1{2^{2n}}}+1\right)}\sim\frac1{2\ln 2}\frac1n$$

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