Does $\sum_{k=2}^{\infty}\left(\ln{\frac{k+1}{k-1}}-\frac{2}{k}\right)$ diverge or converge? 
Does $$\sum_{k=2}^{\infty}\left(\ln{\frac{k+1}{k-1}}-\frac{2}{k}\right)$$ diverge or converge?

It's clear that $f(x)>0 \ \forall \ x\in[2,\infty).$ The function $f(x)=\ln{\frac{x+1}{x-1}}-\frac{2}{x}$ has derivative 
$$f'(x)=-\frac{2}{x^2(x^2-1)}<0 \ \forall \ x \in [2,\infty).$$
So we can use the integral comparison criteria. We have that
$$\int_{2}^{\infty}\left(\ln{\frac{x+1}{x-1}}-\frac{2}{x}\right) dx=\lim_{n\rightarrow\infty}\int_2^n\left(\ln{(1+x)-\ln{(1-x)}}-\frac{2}{x}\right)dx.
$$
I know this integral converges to a fixed value but it's tedious to compute. Are there other comparison tests that would work here?
 A: First the sum should start at $k=2$. By Telescoping sum we have 
$$\sum_{k=2}^{n}\left(\ln{\frac{k+1}{k-1}}-\frac{2}{k}\right) =\color{blue}{\sum_{k=2}^{n}\left(\ln(k+1)-\ln(k)\right)}+\color{red}{\sum_{k=2}^{n}\left(\ln(k)-\ln(k-1)\right)} -\sum_{k=2}^{n}\frac{2}{k} \\= \color{blue}{\ln(n+1)-\ln(2)}+\color{red}{\ln(n)} -\sum_{k=1}^{n}\frac{2}{k}+2\\= 2\left(  \ln(n) -\sum_{k=1}^{n}\frac{1}{k}\right) + 2-\ln2 +\ln{\frac{n+1}{n}} \to -2\gamma +2-\ln2$$
where $\gamma$ is the Euler Mascheroni's constant. That is 
$$\color{blue}{\sum_{k=2}^{\infty}\left(\ln{\frac{k+1}{k-1}}-\frac{2}{k}\right)=-2\gamma +2-\ln2}$$
A: You can write that
$$
\frac{k+1}{k-1}=1+\frac{2}{k-1}\text{ and}\ln\left(1+\frac{1}{k}\right)\underset{(0^{+})}{=}\frac{1}{k}+o\left(\frac{1}{k}\right)
$$
Hence it comes
$$
\ln\left(\frac{k+1}{k-1}\right)-\frac{2}{k}=\frac{2}{k-1}-\frac{2}{k}+o\left(\frac{1}{k-1}\right)=\frac{2}{k\left(k-1\right)}+o\left(\frac{1}{k^2}\right)
$$
Finally
$$
\ln\left(\frac{k+1}{k-1}\right)-\frac{2}{k}\underset{(+\infty)}{\sim}\frac{2}{k^2}$$
Or the series $\displaystyle \sum_{ k \geq 1}^{ }\frac{1}{k^2}$ converges, so as your series.
A: Set $y=x-1\ge 1$. Then
$$
\log\frac{x+1}{x-1}=\log\left(1+\frac{2}{y}\right)=2\left(\frac{1}{y}-\frac{1+o(1)}{y^2}\right).
$$
Therefore
$$
2\sum_{y=1}^\infty \left(\frac{1}{y}-\frac{1+o(1)}{y^2}\right)-\frac{1}{y+1} < \infty.
$$
A: If you use the integral for the logarithm, 
$$\ln{k+1\over k-1}=\int_{k-1}^{k+1}{dx\over x}$$
and interpret the integral as the area beneath the curve $y={1\over x}$, then you can see, from a sketch of the curve, that
$${2\over k}\lt\int_{k-1}^{k+1}{dx\over x}\lt{1\over2}\left({1\over k-1}+{1\over k} \right)+{1\over2}\left({1\over k}+{1\over k+1} \right)$$
hence
$$0\lt\ln\left(k+1\over k-1\right)-{2\over k}\lt{1\over2}\left({1\over k-1}-{1\over k} \right)-{1\over2}\left({1\over k}-{1\over k+1} \right)$$
It follows that
$$0\lt\sum_{k=2}^\infty\left(\ln\left(k+1\over k-1\right)-{2\over k}\right)\lt{1\over2}\sum_{k=2}^\infty\left({1\over k-1}-{1\over k} \right)-{1\over2}\sum_{k=2}^\infty\left({1\over k}-{1\over k+1} \right)={1\over2}\cdot1-{1\over2}\cdot{1\over2}={1\over4}$$
Note, this accords with the exact value in Guy Fsone's answers, $$-2\gamma+2-\ln2\approx-2(0.577)+2-0.693=0.153$$
