|
I have been having problem with calculating the following summation: $$ \sum_{i=1}^n {1\over 4i^2-1} = {1\over3} + {1\over15} + {1\over35} + \cdots + {1\over 4n^2-1} $$ I do know the answer, but just can not find the way to get it. Thank you in advance. |
||||
|
|
|
You have $$ \sum_{i=1}^n {1\over 4i^2-1} = \sum_{i=1}^n\frac{1}{{(2i + 1)(2i - 1)}} $$ $$ = \sum_{i=1}^n \left(\frac{1}{{2(2i - 1)}} - \frac{1}{{2(2i + 1)}}\right) = \frac{1}{2}\left(\sum_{1\leq i\leq n} \frac{1}{{(2i - 1)}} - \sum_{1\leq i\leq n}\frac{1}{{(2i + 1)}}\right) $$ $$ =\frac{1}{2}\left( \frac{1}{1}+\sum_{2\leq i\leq n} \frac{1}{{(2i - 1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right) $$ By translation of index i=k+1 you have $$ =\frac{1}{2}\left( \frac{1}{1}+\sum_{2\leq k+1\leq n} \frac{1}{{(2[k+1] - 1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right). $$ Remember that $2\leq k+1\leq n$ if, onli if, $2-1\leq k\leq n-1$. Then $$ =\frac{1}{2}\left( \frac{1}{1}+\sum_{1\leq k\leq n-1} \frac{1}{{(2k+1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right) $$ $$ =\frac{1}{2}-\frac{1}{4n+2}. $$ |
|||
|
|
|
There is a common trick here: use the fact that $$ \frac{1}{4n^2-1} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right). $$ UPDATE: Also, there's often another way to go about it. If you already know the answer, there is a good chance that you can prove it by induction. It does work in this problem. |
||||
|
|
|
If you use partial fractions, you get a telescoping series: $$\sum_{i=1}^n {1\over 4i^2-1}={1\over 2}\sum_{i=1}^n \Bigl({1\over 2i-1}-{1\over 2i+1}\Bigr)$$ $$={1\over 2}\Bigl(\Bigl(1-{1\over 3}\Bigr)+\Bigl({1\over 3}-{1\over 5}\Bigr)+\cdots +\Bigl({1\over 2n-1}-{1\over 2n+1}\Bigr)\Bigr)$$ $$={1\over 2}\Bigl(1-{1\over 2n+1}\Bigr)={n\over 2n+1}$$ |
|||
|
|
