Convergence/Divergence of an infinite series

Click For Summary
SUMMARY

The series \(\Sigma(\frac{n}{2n+3})^{2}\) can be analyzed for convergence using the Direct Comparison Test. By simplifying the terms, specifically dividing both the numerator and denominator by \(n\), the series can be compared to a known convergent series. The conclusion is that since the terms approach \(\frac{1}{4}\) as \(n\) approaches infinity, the series converges based on the established criteria of the Direct Comparison Test.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with the Direct Comparison Test
  • Basic knowledge of limits and simplification techniques
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the Direct Comparison Test in more detail
  • Learn about the Limit Comparison Test for series
  • Explore p-series and their convergence criteria
  • Practice simplifying complex fractions in series
USEFUL FOR

Students studying calculus, particularly those focusing on series convergence, mathematicians, and educators looking to reinforce concepts related to series analysis.

rdioface
Messages
11
Reaction score
0

Homework Statement


Prove the convergence or divergence of the series \Sigma(\frac{n}{2n+3})^{2} using the Direct Comparison Test.


Homework Equations


If series A converges and every term in series B is less than the corresponding term in series A, then series B converges.
If series C diverges and every term in series D is greater than the corresponding term in series C, then series D diverges.


The Attempt at a Solution


I can't think of anything useful to compare it to. Turning either the top or the bottom to 1 doesn't yield anything useful, nor does it behave as a p-series.
 
Physics news on Phys.org
hi rdioface! :smile:

(use "left(" and "right)" to automatically get latex brackets the right size :wink:)

always try to simplify by getting things as close to 1 as possible …

in this case, try divdiding top and bottom by n :smile:
 

Similar threads

Proving convergence and divergence of series
  • · Replies 3 ·
Replies
3
Views
2K
Valid conclusion for an absolutely convergent sequence
  • · Replies 4 ·
Replies
4
Views
1K
Interval of convergence of power series from ratio test
  • · Replies 2 ·
Replies
2
Views
2K
How to show that these sequences are summable?
  • · Replies 1 ·
Replies
1
Views
2K
Determining if a series converges
  • · Replies 4 ·
Replies
4
Views
2K
Is Using the Comparison Test to Prove Divergence Valid for This Series?
  • · Replies 29 ·
Replies
29
Views
3K
Root test proof using Law of Algebra
  • · Replies 6 ·
Replies
6
Views
2K
Proving the convergence of series
  • · Replies 14 ·
Replies
14
Views
2K
Comparison test of infinite series
  • · Replies 6 ·
Replies
6
Views
2K
Convergence/Divergence of an Infinite Series
  • · Replies 4 ·
Replies
4
Views
2K