Series problem: given s(sub n), find a(sub n) and sum of a(sub n)

[SpicyPepper]

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 \sum a_{n} n=1 to infinity is

s_{n} = (n - 1) / (n + 1)

find a_{n} and \Sigma a_{n} n=1 to infinity

Homework Equations

a_{n} = s_{n} - s_{n-1}

The Attempt at a Solution

I used the equation above to get a_{n} = 2/(n(n+1))

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.

 

[clamtrox]

Not all a_n follow that formula -- check where it's not applicable.

 

[HallsofIvy]

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 \sum a_{n} n=1 to infinity is

s_{n} = (n - 1) / (n + 1)

find a_{n} and \Sigma a_{n} n=1 to infinity

Homework Equations

a_{n} = s_{n} - s_{n-1}

The Attempt at a Solution

I used the equation above to get a_{n} = 2/(n(n+1))

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.

The sum of an infinite series is, by definition, the limit of the sequence of partial sums. Here it is \lim_{n\to\infty} (n-1)/(n+1).

To find the individual terms, a_1 is the first "partial sum": a_1= (1-1)/(1+1)= 0. a_2 is then given by a_1+ a_2= (2-1)/(2+1) so 0+ a_2= a_2= 1/3. a_3 is given by a_1+ a_2+ a_3= 0+ 1/3+ a_3= (3-1)(3+1)= 2/4= 1/2 so a_3= 1/2- 1/3= 1/6, etc.