Question

Use partial fractions to find the sum of the series 1/4n^2-1

Answer 1

Use
\[
\frac{1}{4 n^2-1} =\frac{1}{2 (2 n-1)}-\frac{1}{2
(2 n+1)}
\]
and use telescoping.

Answer 2

Can you remind me of telescoping? Doesnt that mean that most terms will cancel each other out??

Answer 3

yes

Answer 4

Thank you!!

Answer 5

You should find that the sum of the series is
\[\frac 1 2
\]

Answer 6

You have to show your work

Answer 7

First factor the 1/2 out as
\[
\frac 1 2 \left(\frac{1}{2 n-1}-\frac{1}{
2 n+1}\right)
\] and work with
\[
a_n =\frac{1}{2 n-1}-\frac{1}{
2 n+1}
\]
Notice that
\[
a_1 = 1 - \frac 1 3\\
a_2 =\frac 13 -\frac 1 5\\

\]

write
\[ a_3=? -?
\]
then
write
\[a_{n-2}=?-?\\
a_{n-1}=?-?\\
a_{n}=?-?\\


\]
Add all these to get
\[s_n=\sum_{i=1}^n a_i
\]
and deduce that
\[ s_n
\]
converges to 1.
Do not forget the 1/2 we pulled out.

Answer 8

I understand, thanks!!