How do telescoping solutions work in summation?

[karush]

$\tiny(10.r.19)$
I got this answer from W|A

but is this a telescoping solution

\begin{align*}\displaystyle
S_{19}&=\sum_{n=3}^{\infty}\frac{(1)}{(2n-3)(2n-1)}=\frac{1}{6}\\
\end{align*}

 

[MarkFL]

Using partial fraction decomposition on the summand, we may write:

\(\displaystyle S=\frac{1}{2}\sum_{n=3}^{\infty}\left(\frac{1}{2(n-1)-1}-\frac{1}{2n-1}\right)\)

Now, let's re-index...

\(\displaystyle S=\frac{1}{2}\left(\frac{1}{3}+\sum_{n=3}^{\infty}\left(\frac{1}{2n-1}-\frac{1}{2n-1}\right)\right)\)

So yes, this is a telescoping series, and what we're left with is:

\(\displaystyle S=\frac{1}{2}\left(\frac{1}{3}+0\right)=\frac{1}{6}\)

 

[karush]

big Mahalo again ...

I was close to that but got lost again

next week class is over... next fall Calc III :(

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