1. ## partial sums series?

What is the sum of the series 1/ (n x (n+2)) from n=2 as n--> infinity? This series does converge because limit of this is 0. I know that the partial function is (1/2n)- (1/(2(n+2))). Then when I take a look at which way the series go, I end up with S of n = (1/12) - (1/(2(n+2))) - (1/(2(n+2))). I take a limit of (1/12) - lim as n-->infinity of (1/(2(n+2))) - lim as n-->infinity of (1/(2(n+2)))= (1/12)-0-0=(1/12). I don't know where to go after that.

Also how do you find limit from n=1 as n --> infinity of (1/((n)^(1/2)))-(1/((n+1)^(1/2)))? I know this one also converges because limit of this is 0.

Thanks!

2. Originally Posted by shah4u19
What is the sum of the series 1/ (n x (n+2)) from n=2 as n--> infinity? This series does converge because limit of this is 0. I know that the partial function is (1/2n)- (1/(2(n+2))). Then when I take a look at which way the series go, I end up with S of n = (1/12) - (1/(2(n+2))) - (1/(2(n+2))). I take a limit of (1/12) - lim as n-->infinity of (1/(2(n+2))) - lim as n-->infinity of (1/(2(n+2)))= (1/12)-0-0=(1/12). I don't know where to go after that.

Also how do you find limit from n=1 as n --> infinity of (1/((n)^(1/2)))-(1/((n+1)^(1/2)))? I know this one also converges because limit of this is 0.

Thanks!

I supose that "x" in 1/(n x (n+2)) must mean multiplication. Don't ever write it that way without prior explanation: after 7th grade nobody should use "x" as multiplication under otherwise explicitly said. Anyway:

1/[n(n + 2)] = (1/2)[1/n - 1/(n + 2)] =

(1/2)[1/2 - 1/4 + 1/3 - 1/5 +...+ 1/n - 1/( n + 2)] =

(1/2)[1/2 - 1/3 - 1/(n + 1) + 1/(n + 2)] = (1/2)[1/6 - 1/{(n-1)(n+2)}]

Now evaluate the limit of the above expression when n--> oo and you're done...and yes: you get the same as above: 1/12. This is the sum!

Tonio

3. Also how do you find limit from n=1 as n --> infinity of (1/((n)^(1/2)))-(1/((n+1)^(1/2)))? I know this one also converges because limit of this is 0.
Do you mean find the sum ?
$
\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)
$

If so then observe that this is a telescoping series.

$
\sum_{n=1}^{N}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) = 1-\frac{1}{\sqrt{2}} + \left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\dotsb = 1-\frac{1}{\sqrt{N+1}}
$

Now taking the limit

$
\lim\limits_{N\to\infty}\sum_{n=1}^{N}\left(\frac{ 1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) =\lim\limits_{N\to\infty}\left( 1-\frac{1}{\sqrt{N+1}}\right)=1
$

Hope that helps

4. ## Re: partial sums series?

Sorry to revive an old thread. I noticed a minor mistake.

Tonio did most of the hard work in getting the sort-of-telescoping series into the form of

$\displaystyle\frac12\sum_{n=2}^{\infty}\left( \frac{1}{n}-\frac{1}{n+2}\right)$

The only terms that survive the telescoping cancellations are

$\displaystyle\frac12\left(\frac12+\frac13-\frac{1}{n+1}-\frac{1}{n+2}\right)$

Then as $n\to\infty$, the limit goes to $\displaystyle\frac12\left(\frac12+\frac13\right)= \frac12\cdot\frac56=\frac{5}{12}$.