1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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.... $\endgroup$ – user217285 Jul 16 '18 at 21:58
  • $\begingroup$ 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 $\endgroup$ – Clark Makmur Jul 16 '18 at 21:59
2
$\begingroup$

We have that

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

then refer to the Alternating harmonic series and the related Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $.

$\endgroup$
0
$\begingroup$

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 $$1-1/2+1/3-1/4+... = \ln (2) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.