Think about we have a infinite series

summation (j from 0 to infinity) (0.5)^2j=4/3

How did we get this answer?Please help me.

thanks.

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- Jun 17th 2008, 01:16 AMspsInfinite series
Think about we have a infinite series

summation (j from 0 to infinity) (0.5)^2j=4/3

How did we get this answer?Please help me.

thanks. - Jun 17th 2008, 01:20 AMMoo
- Jun 17th 2008, 01:34 AMsps
Thanks Moo

I understood.

I have one more question:

summation (j from 0 to infinity) j*(0.5)^2j=100/225

summation (j from 0 to infinity) j^2*(0.5)^2j=100/135

How did we get those answers? Could you please help me again?

thanks in advance. - Jun 17th 2008, 04:19 AMjjzshen
I can't get line breaks to parse properly on this forum...

There are several ways to do both questions. Here's just an example

View the first one as a geometric series of a geometric series

the second is a little tricker...

Let's try to find a general form for $\displaystyle x + 4x^2 + 9x^3 + \ldots $ assuming |x| is less than 1

there are probably much easier ways to do this - Jun 17th 2008, 05:14 AMcolby2152
Geometric series are of the form $\displaystyle ar^n$, but we have the form $\displaystyle ar^{2n}$, so all we need to do is change the base from $\displaystyle r \rightarrow r^2$

Let $\displaystyle r = 0.5^2$ or $\displaystyle \frac{1}{4}$

We now have the sum, $\displaystyle \sum_0^{\infty} ar^j$, where $\displaystyle a=1,r=0.25$

The sum of a geometric series such as that is $\displaystyle \frac{a}{1-r}$. Therefore, the sum is $\displaystyle \frac{1}{1-0.25} = \frac{1}{0.75} \Rightarrow \frac{1}{\frac{3}{4}} = \frac{4}{3}$. - Jun 17th 2008, 10:58 AMMathstud28
There is a much easier way

Notice that we have a summation of the form

$\displaystyle \sum_{n=0}^{\infty}nx^n$

Where $\displaystyle x=\frac{1}{4}$

Well

$\displaystyle \sum_{n=0}^{\infty}nx^n=x\cdot\sum_{n=1}^{\infty}n x^{n-1}=x\cdot\frac{d}{dx}\bigg[\sum_{n=0}^{\infty}x^n\bigg]$

Now we know that

$\displaystyle \sum_{n=0}^{\infty}x^n=\frac{1}{1-x},\forall{x}\in[-1,1]$

So we have

$\displaystyle x\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]=\frac{x}{(x-1)^2}$

So now we can see that

$\displaystyle \sum_{n=0}^{\infty}n\left(\frac{1}{4}\right)^n=\fr ac{\frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}=\frac{4}{9}$

The second is the same, except you must consider the second derivative