Question with Summing a Series (Non-Geometric)

1. Oct 1, 2009

NastyAccident

1. The problem statement, all variables and given/known data
$$\stackrel{infinity}{n=1}\sum\left(\frac{2}{n^{2}+8n+15}\right)$$

2. Relevant equations
Partial Sums
Knowledge of Series

3. 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:

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

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

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

Sincerely,

NastyAccident

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2. Oct 1, 2009

Office_Shredder

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

3. Oct 1, 2009

HallsofIvy

Staff Emeritus
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?

4. Oct 1, 2009

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.

Sincerely,

NastyAccident