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

In summary: In this problem, ##a_1 = 0## so the first equation becomes ##S_n - S_n-1 = 0##. This is always true and you can solve for ##a_n## by using the second equation.
  • #1
TailRider
2
0

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?
 
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  • #2
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.
 
  • #3
Ahh, I see now; I had a feeling it was something I skipped over. Thanks for the help!
 
Last edited:
  • #4
TailRider said:

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##.
 

1. What does "nth partial sum" mean in the context of a series?

The "nth partial sum" refers to the sum of the first n terms in a series. It is often denoted by Sn.

2. How do you find the value of an in a series when given the nth partial sum?

To find the value of an, also known as the nth term, you can use the formula an = Sn - Sn-1, where Sn is the nth partial sum and Sn-1 is the (n-1)th partial sum.

3. Is there a specific method for finding the sum of a series when given the nth partial sum?

Yes, there are several methods for finding the sum of a series when given the nth partial sum. One common method is to use the formula Sn = (a1 + an)*n/2, where a1 is the first term and an is the nth term.

4. Can the value of an and the sum of a series be found without knowing the nth partial sum?

Yes, there are other methods for finding the value of an and the sum of a series without knowing the nth partial sum. For example, if the series is geometric, you can use the formula Sn = a1(1-rn)/(1-r), where a1 is the first term and r is the common ratio.

5. Are there any special cases or exceptions to keep in mind when finding the value of an and the sum of a series?

Yes, there are some special cases and exceptions to consider when finding the value of an and the sum of a series. For example, if the series is alternating or has a pattern, you may need to use a different formula or approach. It is important to carefully analyze the series and determine the appropriate method to use.

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