Question with Summing a Series (Non-Geometric)

[NastyAccident]

Homework Statement

\stackrel{infinity}{n=1}\sum\left(\frac{2}{n^{2}+8n+15}\right)

Homework Equations

Partial Sums
Knowledge of Series

The Attempt at a Solution

Please see the attached word document for previous work up until this part. Also, please excuse the improper LaTeX usage [I'm getting better]!

To summarize:
\stackrel{infinity}{n=1}\sum\left(\frac{2}{n^{2}+8n+15}\right)
= \stackrel{infinity}{n=1}\sum\left(\frac{1}{n+3}-\frac{1}{n+5}\right)

Now, when I string out the sum I start to have concept issues that neither my book, my lecture notes, nor web searches have been able to explain:
\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{1}{8}\right) + ... + \left(\frac{1}{n}-\frac{1}{n+2}\right) + \left(\frac{1}{n+1}-\frac{1}{n+3}\right) + \left(\frac{1}{n+2}-\frac{1}{n+4}\right) + \left(\frac{1}{n+3}-\frac{1}{n+5}\right)

I think the variable increments are correct at the end, this is one of my questions...

However, another - important - question is which variable increments cancel?

I understand that only:
\left(\frac{1}{4} + \frac{1}{5}\right) + ...

Will remain since those particular terms are lower than the subtracting part of the sum, however, I'm unsure about where to start with canceling variable terms.

Here is my 'guess' on which variable terms will remain:
... + \left(-\frac{1}{n+4}-\frac{1}{n+5}\right)
This was done under the assumption that the positive part of the sum will never reach these particular terms.

From there take the limit of the Series:

lim_{n->infinity} \left(\frac{1}{4} + \frac{1}{5} -\frac{1}{n+4}-\frac{1}{n+5}\right)

lim_{n->infinity} \left(\frac{5}{20} + \frac{4}{20} - 0 - 0 \right)

lim_{n->infinity} \left(\frac{9}{20} \right)




NastyAccident

 

[Office_Shredder]

If you're unsure which terms cancel, write out the first twelve terms and see which ones cancel

 

[HallsofIvy]

Yes, you stopped one term too soon! What is the term immediately after \left(\frac{1}{6}- \frac{1}{8}\right)? So you see now what cancels? When isn+ 3= m+ 5?

 

[NastyAccident]

Ahh, thanks! So, pretty much anything that n+3=n+5 will cancel (on both the beginning and end).

Got it now! Thanks for clearing that blip up.
NastyAccident