Telescoping Sum Convergence: Explained and Solved with Examples | Homework Help

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Homework Statement



The problem asks me to express the sum of the series as a telescoping sum, then find whether it is convergent or divergent. Ok, I get that and how it works and all, but the examples they give in the book are stupid and i on spring break this week so no office hours for professors.

Homework Equations


Here it is:

2/(n^2 + 4n + 3)

I know, easy, but I don't get how to do it...the easy ones stump me.

The Attempt at a Solution



I rewrote it like this:
(1/2)(2/n+3) - 2/n+1)

But the terms do not cancel when I do this. Plus it is an even question so I do not know the solution.
 
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Fuzedmind said:
..but the examples they give in the book are stupid and i on spring break this week

I can tell :smile:

You didn't do the partial fractions right; This is apparent by plugging in n=0 to the original and what you got: \frac{2}{n^2+4n+3}=\frac{1}{n+1}-\frac{1}{n+3}. These terms WILL cancel at some point. Write out the first 5 or so terms of the series and you will see this.
 
Well I did that, and they started cancelling, and I got

(1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) + (1/5 - 1/7)

I canceled the 1/3, 1/4, and the 1/5 out, but where do I go from there?

Sorry I am kind of retarted
 
Now write a few more terms and cancel the 1/6 and 1/7. What terms don't cancel? I kind of have faith that you aren't THAT retarded.
 
What you want to do is make sure that, for each "-1/(n+1)", there exist an m so that its "1/(m+3)" cancels it. That is, given an integer n, what m will make 1/(m+3)= 1/(n+1)?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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