What is the general formula for the sum of a telescoping series?

  • Thread starter Thread starter EstimatedEyes
  • Start date Start date
  • Tags Tags
    Sum
Click For Summary
SUMMARY

The discussion revolves around determining the convergence of the series \(\sum \frac{3}{n(n+3)}\) by expressing it as a telescoping sum. The partial fraction decomposition used is \(\frac{1}{n} - \frac{1}{n+3}\), leading to a partial sum \(s_{n}= (\frac{1}{1} - \frac{1}{4}) + (\frac{1}{2} - \frac{1}{5}) + (\frac{1}{3} - \frac{1}{6}) + ... + (\frac{1}{n} - \frac{1}{n+3})\). The key insight is that the negative term of the nth term cancels with the positive term in the (n+4)th term, resulting in six non-cancelled terms as \(n\) approaches infinity. The series is convergent, and the sum can be derived from the remaining terms.

PREREQUISITES
  • Understanding of telescoping series
  • Knowledge of partial fraction decomposition
  • Familiarity with convergence tests for series
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of telescoping series in detail
  • Learn about convergence tests for infinite series
  • Practice partial fraction decomposition with various functions
  • Explore examples of convergent and divergent series
USEFUL FOR

Students studying calculus, mathematicians interested in series convergence, and educators teaching series and sequences in mathematics.

EstimatedEyes
Messages
31
Reaction score
0

Homework Statement



Determine whether the series is convergent or divergent by expressing s_{n} as a telescoping sum. If it is convergent, find its sum.
\sum\frac{3}{n(n+3)}

Homework Equations


The Attempt at a Solution



Partial Fraction Decomposition: \frac{1}{n} - \frac{1}{n+3}

Partial Sum: s_{n}= (\frac{1}{1} - \frac{1}{4}) + (\frac{1}{2} - \frac{1}{5}) + (\frac{1}{3} - \frac{1}{6}) + (\frac{1}{4} - \frac{1}{7}) + ... + (\frac{1}{n} - \frac{1}{n+3})

From the above partial sum, I deduced that the negative term of the nth term is canceled out by the positive term in the n+4th term. However, from there, I am not able to come up with a general formula for the sum of the series. Any help would be appreciated; thanks!
 
Physics news on Phys.org
So which terms DON'T cancel? I count six of them for large n.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K