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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?

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did you really mean $2^n$ or $2n$? – NoChance Jan 18 '12 at 11:03
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
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
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): $$\eqalign{ \textstyle{1\over 2}\sum\limits_{n=1}^k \Bigl[{(-1)^{n-1} \over 2n+1}+ {(-1)^{n-1} \over 2n-1}\Bigr] &= \textstyle{1\over 2} \Bigl[ (\color{maroon}{1\over3}+1 ) - (\color{darkgreen}{1\over5}+\color{maroon}{1\over3} ) -({1\over7}+\color{darkgreen}{1\over5})+\cdots +(-1)^{k-1}({1\over 2k+1}+{1\over2k-1} ) \Bigr]\cr &= \textstyle{1\over 2} \bigl[1 +(-1)^{k-1} {1\over 2k+1} \bigr]. } $$

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