Does the series $\sum_{n=2}^{\infty}\frac{\sqrt{n^2+1}-n}{\ln(n)}$ converge? 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.
 A: $$\frac{\sqrt{n^2+1}-n}{\ln(n)} = \frac{n \left(\sqrt{1 + \frac{1}{n^2}}-1\right)}{\ln(n)} = \frac{ \frac{1}{2n} + o \left( \frac{1}{n^2}\right)}{\ln(n)} \sim \frac{1}{2 n \ln (n)}$$
so your series is not absolutely convergent.
A: $\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)}
$.
A: 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.
A: By Cauchy condensation test, for the condensed series we obtain
$$\frac{2^n\sqrt{2^{2n}+1}-2^{2n}}{\ln 2^n}=\frac{2^n\sqrt{2^{2n}+1}-2^{2n}}{n\ln 2}\,\frac{2^n\sqrt{2^{2n}+1}+2^{2n}}{2^n\sqrt{2^{2n}+1}+2^{2n}}=$$
$$=\frac{2^{2n}}{n\ln 2(2^n\sqrt{2^{2n}+1}+2^{2n})}=\frac{1}{n\ln 2\left(\sqrt{1+\frac1{2^{2n}}}+1\right)}\sim\frac1{2\ln 2}\frac1n$$
