Infinite Sum of algebraic expression Prove that $$\sum_{i=1}^{\infty} \frac{1}{i(2i+3)} = \frac89 -\frac23\ln2$$
I tried using integration but failed miserably. Hints please.
 A: $$\frac3{2i(2i+3)}=\frac{2i+3-2i}{2i(2i+3)}=\frac1{2i}-\frac1{2i+3}$$
$$\implies I=\sum_{i=1}^{\infty} \frac{1}{i(2i+3)} =\frac23\sum_{i=1}^{\infty}\left(\frac1{2i}-\frac1{2i+3}\right)=\frac23\left(\frac12-\frac15+\frac14-\frac17+\cdots\right)$$
Observe that $1,\dfrac13$ are missing
$$\implies\frac32I=1+\frac13-\left(1-\frac12+\frac13-\frac14-\frac15+\frac16-\frac17+\cdots\right)=\frac43-\ln(1+1)$$
as for $-1<x\le1$ $$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots$$  (See this)
Can you take it home from here?
A: Rewrite
$$
\sum_{i=1}^\infty\frac{1}{i(2i+3)}=\frac23\sum_{i=1}^\infty\left[\frac{1}{2i}-\frac{1}{2i+3}\right].\tag1
$$
Consider geometric progression for $|x|<1$
$$
\sum_{i=1}^\infty x^{i-1}=\frac1{1-x}\tag2
$$
and
$$
\sum_{i=1}^\infty x^{2(i+1)}=\frac{x^4}{1-x^2}.\tag3
$$
Taking integral both of sides $(2)$ and $(3)$ yield
$$
\sum_{i=1}^\infty \frac{x^{i}}i=-\ln(1-x)+C_1\tag4
$$
and
$$
\sum_{i=1}^\infty \frac{x^{2i+3}}{2i+3}=-\frac13x(x^2+3)-\frac12\ln(1-x)+\frac12\ln(x+1)+C_2.\tag5
$$
Set $x=0$ to obtain $C_1$ and $C_2$, then plugging in $(4)$ and $(5)$ to $(1)$ and setting $x=1$ will solve our problem.
A: You can do it by integration too.
$$\begin{aligned}
\frac{2}{3}\sum_{i=1}^{\infty} \left(\frac{1}{2i}-\frac{1}{2i+3}\right) &=\frac{2}{3}\int_0^1 \sum_{i=1}^{\infty} \left(x^{2i-1}-x^{2i+2}\right)\,dx\\
&=\frac{2}{3}\int_0^1 \left(\frac{1}{x}-x^2\right)\frac{x^2}{1-x^2}\,dx=\frac{2}{3}\int_0^1 \frac{(1-x)(1+x+x^2)x}{(1-x)(1+x)}\,dx\\
&=\frac{2}{3}\int_0^1 \left(x+\frac{x^3}{1+x}\right)\,dx=\frac{2}{3}\int_0^1 \left(x+\frac{x^3+1}{x+1}-\frac{1}{x+1}\right)\,dx\\
&=\frac{2}{3}\int_0^1 \left(x+1-x+x^2-\frac{1}{x+1}\right)\,dx \\
&= \boxed{\dfrac{8}{9}-\dfrac{2}{3}\ln 2}
\end{aligned}$$
