MATH220 December 2010
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q8 (a) • Q8 (b) • Q9 (a) • Q9 (b) • Q10 (a) • Q10 (b) • Q10 (c) •
Question 10 (c)

Using (a) and (b) prove that
 $\sum _{n=1}^{\infty }{\frac {1}{4n^{2}1}}={\frac {1}{2}}.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

To show that this sum converges to 1/2, it suffices to show that the limit of the partial sums converges to 1/2.

Hint 2

Use partial fractions to rewrite the summands and you will find a telescoping sum!

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Solution

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Let
 $\displaystyle S_{m}=\sum _{n=1}^{m}{\frac {1}{4n^{2}1}}=\sum _{n=0}^{m}{\frac {1}{(2n1)(2n+1)}}$
To simplify the above sum, we use partial fractions:
$\displaystyle {\frac {1}{(2n1)(2n+1)}}={\frac {A}{2n1}}+{\frac {B}{2n+1}}={\frac {A(2n+1)+B(2n1)}{(2n1)(2n+1)}}$
Thus, $\displaystyle 1=A(2n+1)+B(2n1)$. 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
 $\displaystyle {\begin{aligned}S_{m}&=\sum _{n=1}^{m}{\frac {1}{(2n1)(2n+1)}}\\&={\frac {1}{2}}\sum _{n=1}^{m}\left({\frac {1}{2n1}}{\frac {1}{2n+1}}\right)\end{aligned}}$
Note that this is a telescoping sum! Hence most terms cancel:
 ${\begin{aligned}S_{m}&={\frac {1}{2}}\sum _{n=1}^{m}{\frac {1}{2n1}}{\frac {1}{2}}\sum _{n=1}^{m}{\frac {1}{2n+1}}\\&={\frac {1}{2}}\sum _{k=0}^{m1}{\frac {1}{2(k+1)1}}{\frac {1}{2}}\sum _{n=1}^{m}{\frac {1}{2n+1}}\\&={\frac {1}{2}}\left({\frac {1}{2(0+1)1}}+\sum _{k=1}^{m1}{\frac {1}{2(k+1)1}}\right){\frac {1}{2}}\left(\left(\sum _{n=1}^{m1}{\frac {1}{2n+1}}\right)+{\frac {1}{2m+1}}\right)\\&={\frac {1}{2}}\left(1+\sum _{k=1}^{m1}{\frac {1}{2k+1}}\sum _{n=1}^{m1}{\frac {1}{2n+1}}{\frac {1}{2m+1}}\right)\\&={\frac {1}{2}}\left(1{\frac {1}{2m+1}}\right)\\&={\frac {1}{2}}{\frac {1}{4m+2}}\end{aligned}}$
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.

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