$\sum_{k=1}^{n}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4}$ 
Show that $$\sum_{k=1}^{n}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4}$$
  holds for all $n\in\mathbb{N^+}$.

Since this is a positive series, it suffices to show that $$\sum_{k=1}^{\infty}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4},$$which is true by machine computing. WA gives that 
$$\sum_{k=1}^{\infty}\frac{\ln k}{(2k-1)(2k+1)}\approx 0.231051.$$
 A: By summing by parts,
$$\begin{align}
S_n:&=\sum_{k=1}^{n}\frac{\ln(k)}{(2k-1)(2k+1)}\\
&=\frac{1}{2}\sum_{k=2}^{n}\left(\frac{\ln(k)}{(2k-1)}-\frac{\ln(k)}{(2k+1)}\right)\\
&=\frac{1}{2}\sum_{k=2}^{n}\frac{\ln(k-1)}{(2k-1)}-\frac{1}{2}\sum_{k=2}^{n}\frac{\ln(k)}{(2k+1)}
+\frac{1}{2}\sum_{k=2}^{\infty}\frac{-\ln(1-1/k)}{2k-1}\\
&=-\frac{\ln(n)}{2(2n+1)}+\frac{\ln(2)}{6}+\frac{1}{2}\sum_{k=3}^{\infty}\frac{-\ln(1-1/k)}{2k-1}.
\end{align}$$
Now, by concavity $-\ln(1-x)\leq 3\ln(3/2)x$ for $x\in (0,1/3)$, and therefore
$$S_n< \frac{\ln(2)}{6}+3\ln(3/2)\sum_{k=3}^{\infty}\frac{1}{(2k)(2k-1)}
=\frac{\ln(2)}{6}+3\ln(3/2)\left(\ln(2)-\frac{7}{12}\right)<\frac{1}{4}$$
where we used the fact that $\ln(2)=\sum_{j=1}^{\infty}\frac{(-1)^{j-1}}{j}=\sum_{k=1}^{\infty}\left(\frac{1}{2k-1}-\frac{1}{2k}\right)$.
A: Observe that $$\frac{\ln(k)}{(2k-1)(2k+1)}=\frac{\ln(k)}{4k^2-1}.$$ One may apply the integral test as follows. Note that 
$$\frac {\ln(k)}{4k^2-1}<\frac{100}{99}\cdot\frac{\ln(k)}{4k^2},\forall k>5$$ and
$$\frac{\ln(x)}{4x^2}~{\rm is ~decreasing~,}\forall x\geq 2.$$
It follows that $$\sum_{k=1}^{\infty}\frac{\ln(k)}{4k^2-1}<\sum_{k=1}^5\frac{\ln(k)}{4k^2-1}+\frac{100}{99}\sum_{k=6}^{\infty}\frac{\ln(k)}{4k^2}$$
$$<\sum_{k=1}^5\frac{\ln(k)}{4k^2-1}+\frac{100}{99}\int_5^{\infty}\frac{\ln(x)}{4x^2}~dx <0.25.$$ QED
A: Someone gives a solution as follows, which is essentially equivalent to @Robert Z 's except some details.
\begin{align*} s(n):&=\sum_{k=1}^n\frac{\ln k}{(2k-1)(2k+1)}\\ &\leq\sum_{k=2}^{\infty}\frac{\ln k}{(2k-1)(2k+1)}\\ &=\frac{1}{2}\sum_{k=2}^{\infty}\left(\frac{\ln k}{2k-1}-\frac{\ln k}{2k+1}\right)\\ &=\frac{1}{2}\sum_{k=2}^{\infty}\left(\frac{\ln (k-1)}{2k-1}-\frac{\ln k}{2k+1}\right)+\frac{1}{2}\sum_{k=2}^{\infty}\frac{\ln\left(1+\frac{1}{k-1}\right)}{2k-1}\\ &=\frac{1}{2}\sum_{k=2}^{\infty}\frac{\ln\left(1+\frac{1}{k-1}\right)}{2k-1}\\&=\frac{1}{2}\left(\frac{\ln 2}{3}+\frac{\ln\left(\frac 32\right)}{5}+\sum_{k=4}^{\infty}\frac{\ln\left(1+\frac{1}{k-1}\right)}{2k-1}\right)\\   &\leq\frac{1}{2}\left(\frac{\ln 2}{3}+\frac{\ln \left(\frac{3}{2}\right)}{5}+\sum_{k=4}^{\infty}\frac{1}{(k-1)(2k-1)}\right)\\ &=\frac{1}{2}\left(\frac{\ln 2}{3}+\frac{\ln \left(\frac{3}{2}\right)}{5}+\frac{47}{30}-2\ln 2\right)\\ &<\frac{1}{4}  \end{align*}
