Convergence of a series

  • Thread starter uman
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  • #1
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Hi all. I am stuck on this problem:

Prove that [tex]\sum^{\infty}_{n=1}\frac{n}{(n+1)(n+2)(n+3)}=1/4[/tex].

I am totally stuck on this! I tried breaking it up into partial fractions, which worked on some other problems in this section because it showed how to express them as telescoping series, but in this case that technique provided no new insight. I'm stuck.
 

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  • #2
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What do you get when you use partial fraction decomposition?
 
  • #3
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[tex]\frac{n}{(n+1)(n+2)(n+3)}=-\frac{1}{2(x+1)} + \frac{2}{x+2} - \frac{3}{2(x+3)}[/tex]
 
  • #4
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uman, do you know calculus. to be precise Newton's summation formula? here it is for you.

summation of n/(C + n) where n varies from 0 to Inf is same as

Definite integral of 1 / ( 1 + C.x) from 0 to 1 = 1/C * ln (1+x)

use this formula after you decompose the function into smaller units as you have done.
 
  • #5
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Meaw do you have any links to this formula? I can't find anything on google about it, but I've been at a loss to figure this one out too XD
 
  • #6
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Eh meaw could you give a few more details please?
 
  • #7
HallsofIvy
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There is no such formula. You can approximate a sum by such an integral or use it to determine whether the sum converges or not, but the integral is, in general, NOT equal to the integral.
 
  • #8
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Okay... anyone have any more ideas for solving it?
 
  • #9
Hurkyl
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[tex]\frac{n}{(n+1)(n+2)(n+3)}=-\frac{1}{2(x+1)} + \frac{2}{x+2} - \frac{3}{2(x+3)}[/tex]
Those terms look awfully similar....
 
  • #10
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There is no such formula. You can approximate a sum by such an integral or use it to determine whether the sum converges or not, but the integral is, in general, NOT equal to the integral.
That's a very important point that most people miss. The best example I can think of is,
[tex]\sum_{n=0}^{\infty}\frac{1}{n^2}[/tex]
Which turns out to be [itex]\pi^2/6[/itex]. But the corresponding integral is 1. Proving a is very very difficult, in general...
 

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