How Can You Use Partial Fractions to Find the Sum of an Infinite Series?

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SUMMARY

The discussion focuses on using partial fractions to find the sum of the infinite series Ʃ 4/(n(n+2)) from n=1 to infinity. Participants confirmed that the partial fraction decomposition yields 2/n - 2/(n+2), allowing for cancellation of terms in the series. By writing out the first several terms, users can observe the cancellation that occurs, leading to a clearer understanding of the series' convergence. The key takeaway is that careful manipulation of the series reveals the telescoping nature of the terms.

PREREQUISITES
  • Understanding of partial fraction decomposition
  • Familiarity with infinite series and convergence
  • Basic algebraic manipulation skills
  • Knowledge of telescoping series
NEXT STEPS
  • Study the concept of telescoping series in detail
  • Learn how to apply partial fractions in different contexts
  • Explore convergence tests for infinite series
  • Practice solving similar infinite series problems using partial fractions
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in mastering techniques for summing infinite series using partial fractions.

Sean1218
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Homework Statement



Ʃ 4/(n(n+2)) from n=1 to n=infinity

Homework Equations





The Attempt at a Solution



I tried using partial fractions to get A/n + B/(n+2), and I solved for A and B to get A=2 and B=-2

I tried summing them up, so everything would cancel except the first & last term, but nothing cancels.

I'm not sure of any other methods for finding sums or if I'm not just using this one wrong.

Any help?
 
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Nothing cancels? Yes, partial fractions gives the addend as 2/n- 2/(n+2).
When n= 1, that is 2- 2/3.
When n= 2, that is 1- 2/4.
When n= 3, that is 2/3- 2/5.
When n= 4, that is 2/4- 2/6.
When n= 5, that is 2/5- 2/7.
When n= 6, that is 2/6- 2/8.
When n= 7, that is 2/7- 2/9
etc.

I see a lot cancelling!
 
You have the right idea. Try writing out the first 6 or so terms of$$
\sum_1^\infty(\frac 2 n - \frac 2 {n+2})$$leaving in the parentheses and not simplifying as you go. You will see some terms that cancel in the middle. Once you do that write the first ##n## terms.

[Edit]I see Halls types faster than I do.
 

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