1. ## Infinite series

Think about we have a infinite series
summation (j from 0 to infinity) (0.5)^2j=4/3

thanks.

2. Hello,

Originally Posted by sps
Think about we have a infinite series
summation (j from 0 to infinity) (0.5)^2j=4/3

thanks.
$(0.5)^{2j}=[(0.5)^2]^j=\left(\frac 14\right)^j$

And this is a geometric series

3. 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

4. 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 $x + 4x^2 + 9x^3 + \ldots$ assuming |x| is less than 1

there are probably much easier ways to do this

5. Originally Posted by sps
Think about we have a infinite series
summation (j from 0 to infinity) (0.5)^2j=4/3

thanks.
Geometric series are of the form $ar^n$, but we have the form $ar^{2n}$, so all we need to do is change the base from $r \rightarrow r^2$

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

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

The sum of a geometric series such as that is $\frac{a}{1-r}$. Therefore, the sum is $\frac{1}{1-0.25} = \frac{1}{0.75} \Rightarrow \frac{1}{\frac{3}{4}} = \frac{4}{3}$.

6. Originally Posted by jjzshen
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 $x + 4x^2 + 9x^3 + \ldots$ assuming |x| is less than 1

there are probably much easier ways to do this
There is a much easier way

Notice that we have a summation of the form

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

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

Well

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

Now we know that

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

So we have

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

So now we can see that

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

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