Determinate the sums

May 2010
98
16
Hi all i need to determinate the following sums:
\(\displaystyle \sum_{n=1}^{\infty} (-1)^n\frac{1}{n(n+1)}\)
and
\(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n+1)}\)

How do i approach this? Thanx in advance
 

mr fantastic

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Dec 2007
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Hi all i need to determinate the following sums:
\(\displaystyle \sum_{n=1}^{\infty} (-1)^n\frac{1}{n(n+1)}\)
and
\(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n+1)}\)

How do i approach this? Thanx in advance
Start by substituting the partial fraction decomposition of \(\displaystyle \frac{1}{n(n+1)}\).
 
May 2010
98
16
Start by substituting the partial fraction decomposition of \(\displaystyle \frac{1}{n(n+1)}\).
Well that gives me \(\displaystyle \frac{1}{n} \frac{1}{(n+1)}\)

And then what?:)
 

mr fantastic

MHF Hall of Fame
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Well that gives me \(\displaystyle \frac{1}{n} \frac{1}{(n+1)}\)

And then what?:)
Do you understand what is meant by partial fraction decomposition? \(\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\). I suggest you now write out the first few terms using this and see what happens. It also might help to review your classnotes or textbook. Telescoping series would be a good place to start.
 
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May 2010
98
16
Do you understand what is meant by partial fraction decomposition? \(\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\). I suggest you now write out the first few terms using this and see what happens. It also might help to review your classnotes or textbook. Telescoping series would be a good place to start.
Okay..
I think i found a pattern in the first few terms, as i get

\(\displaystyle 1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +...\) so that means \(\displaystyle 1 - \frac{1}{2} + \frac{1}{2}..\) cancels eachother out meaning the sum is 1.

Then for the alternating series, it gets timed by -1, and by the same idea it sums to -1?
 
May 2010
98
16
Okay..
I think i found a pattern in the first few terms, as i get

\(\displaystyle 1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +...\) so that means \(\displaystyle 1 - \frac{1}{2} + \frac{1}{2}..\) cancels eachother out meaning the sum is 1.

Then for the alternating series, it gets timed by -1, and by the same idea it sums to -1?
EDIT: The alternating series is wrong, gonna calc some more..
 

Prove It

MHF Helper
Aug 2008
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Okay..
I think i found a pattern in the first few terms, as i get

\(\displaystyle 1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +...\) so that means \(\displaystyle 1 - \frac{1}{2} + \frac{1}{2}..\) cancels eachother out meaning the sum is 1.

Then for the alternating series, it gets timed by -1, and by the same idea it sums to -1?
Not all terms are multiplied by -1, only where n is odd... You will have an alternating series that is NOT telescopic.


Have a look at the series

\(\displaystyle \sum_{n = 1}^{\infty}\left[(-1)^n\left(\frac{1}{n} - \frac{1}{n - 1}\right)\right] = -\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) - \left(\frac{1}{3} - \frac{1}{4}\right)\)
\(\displaystyle + \left(\frac{1}{4} - \frac{1}{5}\right) - \left(\frac{1}{5} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{7}\right) - \dots + \dots\)

\(\displaystyle = -1 + \frac{2}{2} - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \frac{2}{6} - \frac{2}{7} + \dots - \dots\)

\(\displaystyle = -1 + \frac{2}{1} - \frac{2}{1} + \frac{2}{2} - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \frac{2}{6} - \frac{2}{7} + \dots - \dots\)

\(\displaystyle = 1 - 2\sum_{n = 1}^{\infty}\left[\frac{(-1)^{n-1}}{n}\right]\)


This infinite sum is the alternating harmonic series, which is known to converge to \(\displaystyle \ln{2}\).


So we finally have

\(\displaystyle \sum_{n = 1}^{\infty}\left[(-1)^n\left(\frac{1}{n} - \frac{1}{n - 1}\right)\right] = 1 - 2\ln{2}\).
 
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