Infinite Sum of algebraic expression

Question

Prove that $$\sum_{i=1}^{\infty} \frac{1}{i(2i+3)} = \frac89 -\frac23\ln2$$

I tried using integration but failed miserably. Hints please.

Answer 1

$$\frac3{2i(2i+3)}=\frac{2i+3-2i}{2i(2i+3)}=\frac1{2i}-\frac1{2i+3}$$

$$\implies I=\sum_{i=1}^{\infty} \frac{1}{i(2i+3)} =\frac23\sum_{i=1}^{\infty}\left(\frac1{2i}-\frac1{2i+3}\right)=\frac23\left(\frac12-\frac15+\frac14-\frac17+\cdots\right)$$

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?

Answer 2

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.

Answer 3

You can do it by integration too.

$$\begin{aligned} \frac{2}{3}\sum_{i=1}^{\infty} \left(\frac{1}{2i}-\frac{1}{2i+3}\right) &=\frac{2}{3}\int_0^1 \sum_{i=1}^{\infty} \left(x^{2i-1}-x^{2i+2}\right)\,dx\\ &=\frac{2}{3}\int_0^1 \left(\frac{1}{x}-x^2\right)\frac{x^2}{1-x^2}\,dx=\frac{2}{3}\int_0^1 \frac{(1-x)(1+x+x^2)x}{(1-x)(1+x)}\,dx\\ &=\frac{2}{3}\int_0^1 \left(x+\frac{x^3}{1+x}\right)\,dx=\frac{2}{3}\int_0^1 \left(x+\frac{x^3+1}{x+1}-\frac{1}{x+1}\right)\,dx\\ &=\frac{2}{3}\int_0^1 \left(x+1-x+x^2-\frac{1}{x+1}\right)\,dx \\ &= \boxed{\dfrac{8}{9}-\dfrac{2}{3}\ln 2} \end{aligned}$$