# Sum of $\frac{1}{1 \cdot 2} + \frac{1}{3\cdot 4} + \cdots$

I am practicing for the GRE, and came across the following question: Find the sum $$\frac{1}{1 \cdot 2} + \frac{1}{3\cdot 4} + \frac{1}{5\cdot 6}\cdots.$$ The answer is given to be $\log(2)$, with the hint: "Apply partial fractions to each term and then recognize the series for $\log(1 + x)$ or estimate." I have no idea what partial fractions means in this context, and have tried without success to manipulate this sum by pulling out terms to get it into a familiar form. Any help is appreciated.

• You're summing terms of the form $$\frac{1}{2k(k-1)},$$ which has partial fraction decomposition $$\frac{1}{2k-1} - \frac{1}{2k}.$$ Thus the series is.... – user217285 Jul 16 '18 at 21:58
• Hint: Taylor series for $$log(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$$, find x that match the above sequence – Clark Makmur Jul 16 '18 at 21:59

## 2 Answers

We have that

$$\sum_{k=1}^n\frac1{(2k-1)2k}=\sum_{k=1}^n \left(\frac1{2k-1}-\frac1{2k} \right)=1-\frac12+\frac13-\frac14+\ldots=\sum_{k=1}^n (-1)^{k+1}\frac1k$$

You have to know two things to solve this problem.

First how to change your fractions to difference of fractions.

For example: $$\frac {1}{(4)(5)}=1/4-1/5$$

Second , you need to know $$1-1/2+1/3-1/4+... = \ln (2)$$