Telescoping Series Simplification

[courtrigrad]

I want to determine whether \sum^{\infty}_{n=2} \frac{2}{n^{2}-1} is convergent or divergent. I did the following:

\sum^{\infty}_{n=2} \frac{2}{n^{2}-1} = \sum^{\infty}_{n=2} (\frac{1}{j-1} + \frac{1}{j+1}). Writing down some terms, I got:

(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5}) + (\frac{1}{4}-\frac{1}{6}).... It seemed like all the terms were canceling except (1+\frac{1}{2}).

But when I looked at the answer, it said the sum was s_{n}= 1+\frac{1}{2}-\frac{1}{n-1}-\frac{1}{n}.

Can anybody tell me how they got this?Thanks

 

[quasar987]

S_n is the partial sum, not the whole sum. to get the whole sum, take the limit of S_n as n-->oo and you do get 1+1/2.

 

[quasar987]

btw, it should be

\sum^{\infty}_{n=2} \frac{2}{n^{2}-1} = \sum^{\infty}_{n=2} (\frac{1}{j-1} - \frac{1}{j+1})

Oh, but you developped the sum correctly, so never mind.

 

[courtrigrad]

but how did they get the partial sum?

 

[quasar987]

but hum... I do calculate something different for S_n... I get

S_n = 1+\frac{1}{2}-\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n+1}

 

[shmoe]

It looks like they want:

s_n=\sum_{j=2}^{n-1}\frac{1}{j-1}-\frac{1}{j+1}

i.e. s_4 is really the first two terms of your sum, that is s_4=(1-1/3)+(1/2-1/4). Then:

s_n=\sum_{j=2}^{n-1}\frac{1}{j-1}-\sum_{j=2}^{n-1}\frac{1}{j+1}=\sum_{j=1}^{n-2}\frac{1}{j}-\sum_{j=3}^{n}\frac{1}{j}

 

[courtrigrad]

So let's say I have \sum_{n=1}^{\infty} (\frac{3}{n(n+3)}) = (\frac{1}{n}-\frac{1}{n+3}). I said that this equals:

(1-\frac{1}{4})+(\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})+(\frac{1}{4}-\frac{1}{7})+...+(\frac{1}{n-1}-\frac{1}{n+2})+(\frac{1}{n}-\frac{1}{n+3}). So how do I simplify this sequence of partial sums so that I can take the limit to get the value of the sumAlso if the nth partial sum of a series \sum_{n=1}^{\infty} a_{n} is s_{n} = 3-\frac{n}{2^{n}} find a_{n} and \sum_{n=1}^{\infty} a_{n}. So would a_{n} = s_{n}-s_{n-1} = (\frac{-n}{2^{n}}+\frac{n-1}{2^{n-1}})? The sum would be \lim_{n\rightarrow \infty} s_{n} = 3?Thanks

 

[shmoe]

So how do I simplify this sequence of partial sums so that I can take the limit to get the value of the sum

The same way I showed you above, break the partial sum into two sums and adjust their variables of summation so the things inside the sums match and you can subtract them (except a few terms on the ends).

So would a_{n} = s_{n}-s_{n-1} = (\frac{-n}{2^{n}}+\frac{n-1}{2^{n-1}})? The sum would be \lim_{n\rightarrow \infty} s_{n} = 3?

Looks fine.