I'm probably in the wrong forum but I have a question on series...

Quote:

Originally Posted by AAKhan07

Hi, I found a question on the front page of the website of my examination board:
MEI - Mathematics in Education and Industry
The 'Maths Item of the Month' is the one you should be looking at.
How do you do this? I think that the answer is found by considering...
$<br /> \sum \frac{1}{n(n+1)}<br />$
But I don't know how to do this sum to infinity, much appreciated...

KHAN

What I see when I look at "Maths Item of the Month" is
$\frac{1}{2^2}+ \frac{1}{2^3}+ \frac{1}{2^4}+\cdot\cdot\cdot+ \frac{1}{2^n}+\cdot\cdot\cdot+ \frac{1}{3^2}+ \frac{1}{3^3}+ \cdot\cdot\cdot+\cdot\dot\cdot+ \frac{1}{4^2}+ /cdot/cdot/cdot$

Each of those, for where the denominators are a powers of the same number is a geometric sequence.
$\frac{1}{a^2}+ \frac{1}{a^3}+\cdot\cdot\cdot= \frac{1}{a^2}\left[1+ \frac{1}{a}+ \frac{1}{a^2}+ \cdot\cdot\cdot$
$= \frac{1}{a^2}\frac{1}{1- 1/a}= \frac{1}{a^2- a}$

In particular, the sum when r= 1/2 is (1/4)(1/(1- 1/2))= 1/2. When r= 1/3, it is (1/9)(1/(1- 1/3))= 1/6. So this sum reduces to $1/2+ 1/3+\cdot\cdot\cdot+ \frac{1}{a^2- a}+ \cdot\cdot\cdot$

$\frac{1}{a^2- a}= \frac{1}{a-1}- \frac{1}{a}$
That means, for example that $\frac{1}{2}= \frac{1}{1}-\frac{1}{2}$ , [tex]\frac{1}{3}= \frac{1}{2}- \frac{1}{3}[\math], $\frac{1}{12}= \frac{1}{3}- \frac{1}{4}$ etc.

In other words, this is a "telescoping" series. The second term in each $\frac{1}{a-1}- \frac{1}{a}$ is canceled by the first term in a second part of the sum.