Proving Convergence of Infinite Series with Changing Signs

In summary, the conversation discusses the convergence of the series \sum_{n=1}^{inf}(\frac{1}{n}-\frac{1}{n+x}) of real-valued functions on R - \{-1, -2, -3, ...\}. The attempt at a solution involves using the telescoping series concept and the comparison test. It is suggested to pick a natural number m such that x<m, and then use the statement of comparison tests to show that the series converges. However, there is a question about the validity of the comparison test when the individual terms can be negative. Another unrelated question is also asked about finding the sum of the series 1/1 + 1/(1+2)
  • #1
ait.abd
26
0

Homework Statement


Show the convergence of the series
[tex] \sum_{n=1}^{inf}(\frac{1}{n}-\frac{1}{n+x})[/tex]
of real-valued functions on [tex] R - \{-1, -2, -3, ...\} [/tex].

Homework Equations


The Attempt at a Solution


I first thought of solving this using telescoping series concept but it didn't work out. Also, I tried to prove it by saying that since denominator goes [tex]n^2[/tex] it should converge but what I doubt is that the elements of the series change sign and the modulus of its terms will not be lower than [tex]1/n^2[/tex].
 
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  • #2
Well, you could try to turn this in a telescoping series.

Pick m a natural number such that x<m. Then

[tex]\frac{1}{n}-\frac{1}{n+x}\leq \frac{1}{n}-\frac{1}{n+m}[/tex]...
 
  • #3
micromass said:
Well, you could try to turn this in a telescoping series.

Pick m a natural number such that x<m. Then

[tex]\frac{1}{n}-\frac{1}{n+x}\leq \frac{1}{n}-\frac{1}{n+m}[/tex]...

Thanks micromass!
 
  • #4
I have a slight problem with the solution micromass hope you can clarify. I was looking at the statement of comparison tests that states that [tex]a_n, b_n > 0 [/tex] for comparison test to be valid whereas in the solution above we can have negative individual terms as well? Is it so or I am looking at it wrongly?
e.g.
[tex]n=1, x=-0.1 \implies 1-10/9 < 0[/tex]
 
  • #5
Given the series:

1/1 + 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) + ...

How would u find the sum of the series?
Any help pleaseee!
 

1. What is the sum of an infinite series?

The sum of an infinite series is the result obtained when all the terms in the series are added together. This sum can be a finite number, infinite, or may not exist at all.

2. How do you calculate the sum of an infinite series?

The sum of an infinite series can be calculated by using a mathematical formula, such as the geometric series formula or the telescoping series formula. Alternatively, it can also be calculated by taking the limit of the partial sums of the series as the number of terms approaches infinity.

3. What is the difference between a convergent and a divergent infinite series?

A convergent infinite series is one whose sum approaches a finite number as the number of terms increases, while a divergent infinite series is one whose sum approaches infinity or does not exist at all.

4. How do you determine if an infinite series is convergent or divergent?

There are several tests that can be used to determine if an infinite series is convergent or divergent, such as the ratio test, the root test, and the integral test. These tests involve analyzing the behavior of the terms in the series to determine if they approach zero or a finite value as the number of terms increases.

5. What are some real-world applications of infinite series?

Infinite series have various applications in fields such as physics, engineering, and economics. For example, they can be used to model the growth of populations, the behavior of electrical circuits, and the value of investments over time. They also play a significant role in calculus and other branches of mathematics.

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