Science:Math Exam Resources/Courses/MATH220/December 2010/Question 10 (c)/Solution 1

Let

$\displaystyle S_{m}=\sum _{n=1}^{m}{\frac {1}{4n^{2}-1}}=\sum _{n=1}^{m}{\frac {1}{(2n-1)(2n+1)}}$

To simplify the above sum, we use partial fractions:

$\displaystyle {\frac {1}{(2n-1)(2n+1)}}={\frac {A}{2n-1}}+{\frac {B}{2n+1}}={\frac {A(2n+1)+B(2n-1)}{(2n-1)(2n+1)}}$

Thus, $\displaystyle 1=A(2n+1)+B(2n-1)$. Plugging in $\displaystyle n=1/2$ shows that $\displaystyle A=1/2$ and plugging in $\displaystyle n=-1/2$ shows that $\displaystyle B=-1/2$. Hence

$\displaystyle {\begin{aligned}S_{m}&=\sum _{n=1}^{m}{\frac {1}{(2n-1)(2n+1)}}\\&={\frac {1}{2}}\sum _{n=1}^{m}\left({\frac {1}{2n-1}}-{\frac {1}{2n+1}}\right)\end{aligned}}$

Note that this is a telescoping sum! Hence most terms cancel:

${\begin{aligned}S_{m}&={\frac {1}{2}}\sum _{n=1}^{m}{\frac {1}{2n-1}}-{\frac {1}{2}}\sum _{n=1}^{m}{\frac {1}{2n+1}}\\&={\frac {1}{2}}\sum _{k=0}^{m-1}{\frac {1}{2(k+1)-1}}-{\frac {1}{2}}\sum _{n=1}^{m}{\frac {1}{2n+1}}\\&={\frac {1}{2}}\left({\frac {1}{2(0+1)-1}}+\sum _{k=1}^{m-1}{\frac {1}{2(k+1)-1}}\right)-{\frac {1}{2}}\left(\left(\sum _{n=1}^{m-1}{\frac {1}{2n+1}}\right)+{\frac {1}{2m+1}}\right)\\&={\frac {1}{2}}\left(1+\sum _{k=1}^{m-1}{\frac {1}{2k+1}}-\sum _{n=1}^{m-1}{\frac {1}{2n+1}}-{\frac {1}{2m+1}}\right)\\&={\frac {1}{2}}\left(1-{\frac {1}{2m+1}}\right)\\&={\frac {1}{2}}-{\frac {1}{4m+2}}\end{aligned}}$

Now, the question asks us to evaluate $\displaystyle \lim _{m\to \infty }S_{m}$. Using parts (a) and (b), we have that

$\displaystyle \lim _{m\to \infty }S_{m}=\lim _{m\to \infty }\left({\frac {1}{2}}-{\frac {1}{4m+2}}\right)=1/2$

completing the question.