Partial Sum of a Series (k-1)/(k+1)

[runningeagle]

"Suppose the kth partial sum of a series a[sub:2f0xv54c]n[/sub:2f0xv54c] from 1 to infinity is given by s[sub:2f0xv54c]k[/sub:2f0xv54c]=(k-1)/(k+1) for k?1.

a)Find a formula for the general term a[sub:2f0xv54c]n[/sub:2f0xv54c] for n ?2.

b)Does the series a[sub:2f0xv54c]n[/sub:2f0xv54c] from 1 to infinity converge? Find its sum."

So, for part a, I have a[sub:2f0xv54c]n[/sub:2f0xv54c]=[1/3+2/4+3/5+4/6+...]. I have the general term is (n-1)/(n+1).

For part b, I have the series does not converge because of the limit test (the sequence does not converge).

Am I doing this correctly? The part about the partial sums is throwing me off.

 

[tkhunny]

You seem to be mising up a few things.

a(n) are the terms.

s(k) are the partial sums.

For the kth term, you must subtract successive sums. Find and simplify s(k) - s(k-1) since that is a(k).

Let's see what you get.

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ S_n \ = \ \frac{n-1}{n+1} \ = \ 1-\frac{2}{n+1}\)

\(\displaystyle Find \ (a) \ a_n, \ n \ \ge \ 2, \ and \ (b), \ if \ a_n \ converges, \ its \ sum.\)

\(\displaystyle Ergo, \ a_n \ = \ S_n-S_{n-1} \ = \ \bigg(1-\frac{2}{n+1}\bigg)-\bigg(1-\frac{2}{n}\bigg) \ = \ \frac{2}{n}-\frac{2}{n+1} \ = \ \frac{2}{n(n+1)}\)

\(\displaystyle (a) \ Therefore, \ a_n \ = \ \sum_{n=2}^{\infty}\frac{2}{n(n+1)} \ = \ 1-\frac{2}{n+1}\)

\(\displaystyle (b) \ \lim_{n\to\infty}S_n \ = \ \lim_{n\to\infty}\bigg(1-\frac{2}{n+1}\bigg) \ = \ 1, \ hence, \ series \ converges \ to \ 1.\)

 

[runningeagle]

Thanks!
Could you explain why (n-1)/(n+1) =1-2(n+1)?

I tried multiplying by the conjugate (n+1)/(n+1), but did not get 1-2(n+1)?

 

[BigGlenntheHeavy]

\(\displaystyle \frac{n-1}{n+1} \ = \ 1-\frac{2}{n+1} \ not \ 1-2(n+1)\)