Sum of infinite series

  • Thread starter ait.abd
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  • #1
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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].
 

Answers and Replies

  • #2
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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
26
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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
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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
8
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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!!
 

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