Given nth partial sum of a series, find a of n and sum

[TailRider]

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?

 

[Fightfish]

The problem here is that $a_{n} = \frac{2}{n(n+1)}$ only for $n\neq 1$. By using $a_{n} = S_{n} - S_{n-1}$, you made the implicit assumption that $S_{n-1}$ exists, which for $n = 1$, it clearly does not.

 

[TailRider]

Ahh, I see now; I had a feeling it was something I skipped over. Thanks for the help!

 

[Ray Vickson]

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?

The nth partial sum means $S_n = \sum_{k=1}^n a_k$, so $S_1 = a_1$ and $S_n = a_n + S_{n-1}$ for $n > 1$.