Nth partial sum of a series Σ Ak is Sn = (n-1)/(n+1)

[Kenshin]

If the nth partial sum of a series Σ Ak is Sn = (n-1)/(n+1) , find Ak . Does Σ Ak converge?

i looked in my math book and can't find how to do this.

 

[Tide]

Try subtracting S(n) from S(n+1).

 

[HallsofIvy]

Either I or Tide misunderstands your question. Subtracting S_n from S_n+1 will give the A_n+1 but won't tell you whether or not the sequence converges.

The DEFINITION of convergence of a series is the convergence of the sequence of partial sums.

Can you find the limit of the sequence $\frac{n-1}{n+1}$ as n goes to infinity??

 

[Kenshin]

Ak= 2/x^2+3x+2. so you subtrack Sn+1-Sn. can u show me how that works. it just seems weird that, that is all u have to do. thanks

 

[TenaliRaman]

Kenshin,
Look at the last statement made by HallsOfIvy ...
Have u worked out what he said?

-- AI

 

[shmoe]

Ak= 2/x^2+3x+2. so you subtrack Sn+1-Sn. can u show me how that works. it just seems weird that, that is all u have to do. thanks

To find $$A_k$$ yes, this is all you have to do (you replaced your "k" with "x" though, and also "n"). Do you understand why this works?

To answer the question of convergence for the series, see HallsofIvy's post. You have to look at $$\lim_{n\rightarrow\infty}S_n$$

 

[matt grime]

Ak= 2/x^2+3x+2. so you subtrack Sn+1-Sn. can u show me how that works. it just seems weird that, that is all u have to do. thanks

Let A(k) be any series, and S(n) the sum of the first n terms, then S(n+1) is the sum of the first n terms plus the n+1'st term. Can you now prove the result that puzzles you?

 

[HallsofIvy]

I DID misunderstand the question! There were TWO things asked:

1) Find A(k) which is what Tide was responding to.

Since S(n), the "partial sum is defined as A(1)+ A(2)+ ...+ A(n),
A(n)= S(n)- S(n-1)= (n-1)/(n+1) - (n-1-1)/(n-1+1)= (n-1)/(n+1)- (n-2)/n
= n(n-1)/n(n+1)- (n+1)(n-2)/n(n+1)= (n²-n-n²+n+2)/n(n+1)
= 2/n(n+1).

2) Since A(k) converges if and only if the sequence of partial sums converges, look at
S(n)= (n-1)/(n+1). Divide both numerator and denominator by n: (1- 1/n)/(1+ 1/n).
As n goes to infinity, 1/n goes to 0 so the S(n) converges to 1.

Yes, Σ A(k) converges. In fact, it converges to 1.

 

[matt grime]

I don't think you mean if and only if in part 2) there.

 

[HallsofIvy]

I don't think you mean if and only if in part 2) there.

? That's certainly the definition of "convergence of a series" that I learned.

If the sequence of partial sums converges, then the series converges to the same limit.

If the sequence of partial sums does not converge, then the series does not converge.

 

[shmoe]

2) Since Σ A(k) converges if and only if the sequence of partial sums converges,

I inserted missing Σ?

 

[matt grime]

The reason I didn't put a correction is that two spring to mind. A(k) converges if the sequence of partial sums converges, or by definition the sum of the sequence converges if and only if the sequence of partial sums converges, and I wasn't sure which Hallsofivy meant. It seems from your reply, that you intended the second option.

 

[HallsofIvy]

Oh- I forgot the Σ!