# Evaluate the sum of series

Find the sum of following series:

$\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...$

After some arrangement, I got below step:

$\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...=\sum\limits_{k=0}^{\infty}\frac{1}{2k+2}- \sum\limits_{k=0}^{\infty}\frac{1}{2k+3}$

Now, I have no idea how to find this difference of two series. Please help me. Thanks in advance.

• Expand $$\ln(1+x)$$ – lab bhattacharjee Dec 29 '15 at 15:45
• The series on the left converges, and the two series on the right do not, so that approach may not be useful. Your series is $\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots$. Now have you seen something like this before? – André Nicolas Dec 29 '15 at 15:53
• @lab battacharjee $1-\log 2$ – Siddhant Trivedi Dec 29 '15 at 15:55

We have $$\sum_{k=1}^{\infty}\frac1{2k(2k+1)}=\sum_{k=1}^{\infty}\left(\frac1{2k}-\frac1{2k+1}\right)=\sum_{n=2}^{\infty}\frac{(-1)^n}{n}=1+\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=1-\ln2.$$ The last step follows from $\ln (1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}$ by setting $x=-1$.
• @zz20s The $k$th term in the first expression is the sum of $2k$th and $(2k+1)$th terms in the second (e.g. $\frac12-\frac13=\frac{(-1)^2}2 +\frac{(-1)^3}3$). – Start wearing purple Dec 29 '15 at 16:39