- #1
SpicyPepper
- 20
- 0
Homework Statement
This is a problem in my book, and the answer is in the back. Unfortunately I can't solve it.
If the nth partial sum of a series [tex]\sum a_{n}[/tex] n=1 to infinity is
[tex]s_{n} = (n - 1) / (n + 1)[/tex]
find [tex]a_{n}[/tex] and [tex]\Sigma a_{n}[/tex] n=1 to infinity
Homework Equations
[tex]a_{n} = s_{n} - s_{n-1}[/tex]
The Attempt at a Solution
I used the equation above to get [tex]a_{n} = 2/(n(n+1))[/tex]
To get the sum, I wrote out the sequence after I figured out:
2/(n(n+1)) = 2/n - 2/(n+1)
so
= lim n->infinity (2/1 - 2/2) + (2/2 - 2/3) + (2/3 - 2/4) + ... + (2/n - 2/(n+1))
= lim n->infty 2/1 - 2(n+1)
= 2
But the answer for the sum is 1.