$\sum_{k=1}^{n}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4}$ 
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 
$$\sum_{k=1}^{\infty}\frac{\ln k}{(2k-1)(2k+1)}\approx 0.231051.$$

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
