- #1
TailRider
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Homework Statement
If the nth partial sum of a series ##\sum_{n=1} ^\infty a_{n}## is
##S_{n} = \frac {n-1} {n+1}##
Find ##a_{n}## and ##\sum_{n=1}^\infty a_n##
Homework Equations
##S_{n} - S_{n-1}= a_{n}##
##\lim_{n \rightarrow +\infty} {S_{n}} = \sum_{n=1}^\infty a_n = S##
The Attempt at a Solution
So I used the first equation and I found ##a_{n} = \frac {2} {n(n+1)}##
Work: ##S_{n} - S_{n-1}= a_{n}= \frac {n-1} {n+1} - \frac {n-2} {n} = \frac {(n-1)(n) - (n+1)(n-2)} {n(n+1)} = \frac {n^2 -n -n^2 - (-n -2)}{n(n+1)} = \frac {2} {n(n+1)}##
The part I'm having a hard time understanding is the sum part.
##\lim_{n \rightarrow +\infty} {\frac {n-1} {n+1}} = 1##, however, ##\sum_{n=1}^\infty \frac {2} {n(n+1)} = 2(\sum_{n=1}^\infty \frac {1} {n(n+1)})##, and ##\sum_{n=1}^\infty \frac {1} {n(n+1)}## is a telescoping series that comes out as 1, so shouldn't the sum of the series equal 2? Why do these come out as 2 different answers even though they are suppose to be the same?
The answer at the back of the book says ##a_{1} = 0##, and I have a hard time understanding this, as when n=1, ##a_{1} = 2/2 = 1##. If it's because you plug 1 into ##S_{n}## and you get 0, then that still leaves me confused, as ##S_{1} = a_{1}##, but it isn't, or is my understanding incorrect?