Sequences + Integrate + Summation

$\Large a_n=\dfrac{1}{x}\huge |_{\small{\frac{n(n+2)}{2}}}^{\small n}$ $=\dfrac{1}{n}\left(1-\dfrac{2}{n+2}\right)$ $\Large \implies a_n=\dfrac{1}{n+2}$ $\begin{aligned}\displaystyle\sum_{n=1}^{\infty}\frac{a_n}n&=\dfrac 12\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)\\&\\&\mathbf{A~ Telescopic ~Series} \\&\\&=\dfrac 12\left(1+\dfrac 12\right)=\dfrac 34=\boxed{0.75}\end{aligned}$