Convergence test of $S=\frac{1}{\ln 2} \sum_{k=1}^\infty \ln (1+\frac{1}{k(k+2)}) \ln k$ Does S converge? (The answer says it converges)

$S=\frac{1}{\ln 2} \sum_{k=1}^\infty \ln (1+\frac{1}{k(k+2)}) \ln k$

My attempt:

Comparison test: $\ln (1+\frac{1}{k(k+2)}) \ln k \lt \ln 2 \ln k$ --- not work;
Cauchy's test: $\lim_{k \to \infty} |\ln (1+\frac{1}{k(k+2)}) \ln k|^{1/k} \lt 1$? --- hard to tell
Ratio test: $\lim_{k \to \infty} |\frac{\ln (1+\frac{1}{(k+1)(k+3)})\ln (k+1)}{\ln (1+\frac{1}{k(k+2)})\ln k} | \lt 1$? --- hard to tell
Integral test: is the series decreasing and founded?

Thanks,
 A: $\ln(1+x)\lt x$ for $x\to 0$ so the sum is, term by term, smaller than
$$\dots \lt \frac{1}{\ln 2} \sum_{k=1}^{\infty} \frac{\ln k}{k(k+2)} $$
but this is clearly convergent because the logarithm is just a small correction relatively to the power laws, and $\sum 1/k^2$ safely converges. Alternatively, we may use e.g. $\ln k < \sqrt{k+100}$ for $k\geq 1$ so the latest sum is smaller than
$$\dots \lt\frac{1}{\ln 2} \sum_{k=1}^\infty \frac{\sqrt{k+100}}{k(k+2)}  $$
but this is already seen to be as convergent as $\sum 1/k^{3/2}$ i.e. as $\int 1/k^{3/2}$ whose indefinite integral is $\sim k^{-1/2}\to 0$ for $k\to\infty$.
A: You should use the equivalence
$$\ln(1+x)\sim_{x\to 0} x$$
Here as $k$ goes to $\infty$, you get
$$\ln(1+\frac{1}{k(k+2)})\ln(k)\sim \frac{\ln(k)}{k(k+2)}\sim \frac{\ln(k)}{k^2}.$$
Then, either you know the criterion of convergence of the serie $\sum_k\ln^a(k)k^b$, or you reprove it in this particular case, noticing for example that
there is a constant $M$ such that
$$\forall k\in\mathbb{N}^*, \left|\frac{\ln(k)}{k^2}\right|\leq \frac{M}{k^{3/2}}.$$
A: Note that $\log(1+x)<0$ for $x\to 0$ therefore $$\frac{1}{\ln 2} \sum_{k=1}^\infty \ln (1+\frac{1}{k(k+2)}) \ln k<\frac{1}{\ln 2} \sum_{k=1}^\infty \frac{1}{k(k+2)} \ln k$$ using the Cauchy condensation test $$\frac{1}{\ln 2} \sum_{k=1}^\infty  2^k\frac{1}{2^k(2^k+2)} k\ln 2=\frac{1}{\ln 2} \sum_{k=1}^\infty  \frac{1}{(2^k+2)} k\ln 2$$ but $$\frac{1}{(2^k+2)} k\ln 2\sim_{k\to +\infty} \frac{1}{(2^k)} k\ln 2$$ and the series  $$\sum_{k=1}^\infty  \frac{1}{(2^k)} k\ln 2$$ converges. Indeed using the root test $$\lim_{k\to +\infty} \frac{1}{({2^k}^{\frac{1}{k}})} k^{\frac{1}{k}}=\frac {1}{2}<1$$
