nth partial sum

Apr 2009
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0
Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

1)



I have no idea how to do this one. I missed it on the test and it might show up on the final
 

chisigma

MHF Hall of Honor
Mar 2009
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near Piacenza (Italy)
The 'key' is the identity...

\(\displaystyle \frac{1}{(n+1)(n+2)}= \frac{1}{n+1} - \frac{1}{n+2}\) (1)

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Feb 2010
1,036
386
Dirty South
Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

1)



I have no idea how to do this one. I missed it on the test and it might show up on the final
\(\displaystyle \frac{7}{(n+1)(n+2)} = \frac{7}{n+1}-\frac{7}{n+2}\)

Now,

\(\displaystyle \sum_{n=1}^k \frac{7}{n+1}-\frac{7}{n+2} = \left( \frac{7}{2}-\frac{7}{3} \right) + \left( \frac{7}{3}-\frac{7}{4} \right)+....+ \left( \frac{7}{k+1}-\frac{7}{k+2}\right)\)

\(\displaystyle = \left( \frac{7}{2} -\frac{7}{k+2}\right)\)

find the limit as k tends to infinity.
 
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Prove It

MHF Helper
Aug 2008
12,897
5,001
Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

1)



I have no idea how to do this one. I missed it on the test and it might show up on the final
Surely this is

\(\displaystyle 7\sum_{n = 2}^{\infty}\frac{1}{n(n + 1)}\).


Using partial fractions:

\(\displaystyle \frac{A}{n} + \frac{B}{n + 1} = \frac{1}{n(n + 1)}\)

\(\displaystyle \frac{A(n + 1) + Bn}{n(n + 1)} = \frac{1}{n(n + 1)}\)

\(\displaystyle A(n + 1) + Bn = 1\)

\(\displaystyle (A + B)n + A= 0n + 1\).

So \(\displaystyle A = 1\) and \(\displaystyle B = -1\).


Therefore we can write the sum as

\(\displaystyle 7\sum_{n = 2}^{\infty}\frac{1}{n(n + 1)} = 7\sum_{n = 2}^{\infty}\left(\frac{1}{n} - \frac{1}{n + 1}\right)\)

This is a telescoping series, so all terms except the first and last will cancel.

So the \(\displaystyle n^{\textrm{th}}\) partial sum is

\(\displaystyle 7\left(\frac{1}{2} - \frac{1}{n + 1}\right)\).


Therefore, the sum of the series is

\(\displaystyle \lim_{n \to \infty}7\left(\frac{1}{2} - \frac{1}{n + 1}\right)\)

\(\displaystyle = 7\left(\frac{1}{2} - 0\right)\)

\(\displaystyle = \frac{7}{2}\).
 
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