Prove that a series converges.

  • Thread starter anoncow
  • Start date
  • Tags
    Series
In summary, the conversation discusses the convergence of the series $\sum\limits_{n=1}^\infty \frac{1 - (-1)^n e}{1 + (n \pi)^2}$. The experts suggest using the comparison test with $\frac{1}{n^2}$ as the comparison series. It is then shown that the series absolutely converges.
  • #1
anoncow
4
0

Homework Statement


$$ \sum\limits_{n=1}^\infty \frac{1 - (-1)^n e}{1 + (n \pi)^2}$$

Homework Equations



The Attempt at a Solution


I'd imagine you have to use the comparison test on it but am unable to figure out where to start. Any suggestions would be greatly appreciated.
 
Last edited:
Physics news on Phys.org
  • #2
Well indeed the comparison test will do (with 1/n^2).

I'll let you see how exactly.
 
  • #3
Interesting sum, compare to 1/n^2
 
  • #4
Is this sufficient proof?

http://mathb.in/1308?key=f6d1a50cb6e98de9513ed7740481e9387911e3ce
 
Last edited:
  • #5
Much simpler than that:

[tex]|\frac{1-(-1)^ne}{1+(\pi n)^2}| \leq \frac{1+e}{1+(\pi n)^2}[/tex]
 
  • #6
MathematicalPhysicist said:
Much simpler than that:

[tex]|\frac{1-(-1)^ne}{1+(\pi n)^2}| \leq \frac{1+e}{1+(\pi n)^2}[/tex]

oh so could i just have done:

[tex]0 \leq |\frac{1-(-1)^ne}{1+(\pi n)^2}| \leq |\frac{1}{n^2}|[/tex]
[tex]\therefore \frac{1-(-1)^ne}{1+(\pi n)^2}[/tex] absolutely converges.
 

1. What does it mean for a series to converge?

A series converges if the sum of its terms approaches a finite value as the number of terms increases towards infinity. In other words, the series "settles down" to a specific value as more and more terms are added.

2. How do you prove that a series converges?

There are several methods for proving that a series converges, including the comparison test, the ratio test, and the root test. These tests involve comparing the series to other known convergent or divergent series and using mathematical manipulations to determine the convergence or divergence of the original series.

3. Can a series converge to more than one value?

No, a series can only converge to a single value. If a series converges, it means that the terms are "settling down" to a specific value and will not change. If a series appears to converge to multiple values, it is likely that the series is divergent.

4. What happens if a series does not converge?

If a series does not converge, it is said to be divergent. This means that the sum of the terms does not approach a finite value as the number of terms increases towards infinity. Divergent series can either have an infinite sum or oscillate between different values.

5. Can you give an example of a convergent series?

One example of a convergent series is the geometric series 1/2 + 1/4 + 1/8 + 1/16 + ..., which converges to a sum of 1. This can be proven using the ratio test, as the ratio between consecutive terms approaches 1/2 as the number of terms increases towards infinity.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
160
  • Calculus and Beyond Homework Help
Replies
1
Views
217
  • Calculus and Beyond Homework Help
Replies
1
Views
181
  • Calculus and Beyond Homework Help
Replies
4
Views
263
  • Calculus and Beyond Homework Help
Replies
3
Views
397
  • Calculus and Beyond Homework Help
Replies
2
Views
696
  • Calculus and Beyond Homework Help
Replies
6
Views
461
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
207
  • Calculus and Beyond Homework Help
Replies
5
Views
471
Back
Top