How to evaluate what a series converges to?

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Homework Help Overview

The discussion revolves around evaluating the convergence of the series \(\sum_{n=1}^\infty \frac{1}{n(n+1)}\). Participants explore the use of partial fractions and the concept of partial sums in relation to this series.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of partial fractions to rewrite the series and question how to evaluate the resulting expression. There is mention of changing variables in the summation and regrouping terms to identify cancellations.

Discussion Status

The conversation is ongoing, with participants offering insights into the evaluation process and questioning the definition of the partial sum \(S_n\). Some guidance is provided regarding the relationship between partial sums and the infinite series.

Contextual Notes

There is uncertainty regarding the appropriate form of \(S_n\) and how to approach the limit as \(n\) approaches infinity. Participants are also navigating the implications of changing variables in the summation.

Chandasouk
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I was asked to evaluate the summation of [tex]\frac{1}{n(n+1)}[/tex] from n=1 to infinity

I used partial fractions to obtain [tex]\frac{1}{n}[/tex] - [tex]\frac{1}{n+1}[/tex]

From here I don't understand how to evaluate. In my solutions manual, they plugged in values from 1 to infinity showing (1 - 1/2+ (1/2 - 1/3)...etc and created a new series called Sn = 1 - [tex]\frac{1}{n+1}[/tex] then took the limit of that to infinity to get the answer 1.

How would I know what Sn should be?
 
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So you want to evaluate:

[tex]\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \frac{1}{n} - \sum_{n=1}^\infty \frac{1}{n+1}[/tex]

One way to evaluate it is to consider what happens if you change variables in the second sum to m = n + 1. If you make this change of variables, do you see how you can deduce the solution?
 
Your series partial sum is (1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1)). Regroup that like this 1+(-1/2+1/2)+(-1/3+1/3)+...+(-1/n+1/n)-1/(n+1). Lots of stuff cancels.
 
Chandasouk said:
How would I know what Sn should be?
Quick check -- you realize they are (presumably) using Sn for the n-th partial sum? And remember that an infinite sum is the limit of the partial sums?
 

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