Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Which is the easiest way to find the value of $$\sum \limits_{n=1}^k \frac{(-1)^{n-1}\times 2n}{{(2n-1)\times (2n+1)}} $$

There is a hint, which says we can use $$ {{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} } $$

but I am not sure how, any ideas guys?

share|improve this question
3  
did you really mean $2^n$ or $2n$? –  Emmad Kareem Jan 18 '12 at 11:03
3  
In case you did mean $2n$ instead of $2^n$ try substituting $r=2n$, otherwise the hint doesn't make much sense. –  E.O. Jan 18 '12 at 11:17
1  
Like in the previous comments, I am assuming there is a typo, it should be $2n$. Using the hint, write down the sum of the first $4$ terms. –  André Nicolas Jan 18 '12 at 14:20
    
and note it collapses... –  David Mitra Jan 18 '12 at 15:13
    
Yes, it was a typo, @Andre post your hint as an answer, I want to accept it. –  Quixotic Jan 18 '12 at 15:33

1 Answer 1

up vote 2 down vote accepted

From the hint, use partial fractions to write $$ {(-1)^{n-1}2n\over (2n-1)(2n+1) }={1\over2}{(-1)^{n-1}\over 2n+1}+{1\over2}{(-1)^{n-1}\over 2n-1} $$ So, $$ \sum_{n=1}^k {(-1)^n 2n\over(2n-1)(2n+1) }= {1\over 2}\sum\limits_{n=1}^k \Bigl[{(-1)^{n-1} \over 2n+1}+ {(-1)^{n-1} \over 2n-1}\Bigr]. $$ The series is collapsing (the left hand term in one parenthetical expression below cancels with the right hand term in the next parenthetical expression): $$\eqalign{ \textstyle{1\over 2}\sum\limits_{n=1}^k \Bigl[{(-1)^{n-1} \over 2n+1}+ {(-1)^{n-1} \over 2n-1}\Bigr] &= \textstyle{1\over 2} \Bigl[ (\color{maroon}{1\over3}+1 ) - (\color{darkgreen}{1\over5}+\color{maroon}{1\over3} ) -({1\over7}+\color{darkgreen}{1\over5})+\cdots +(-1)^{k-1}({1\over 2k+1}+{1\over2k-1} ) \Bigr]\cr &= \textstyle{1\over 2} \bigl[1 +(-1)^{k-1} {1\over 2k+1} \bigr]. } $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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