# Limit of a series

1. Apr 4, 2006

### sweetvirgogirl

so there is a power series
S 4^n z^(3n)
and upper limit being infinity and lower limit being 0. (S means sigma)
then my book says, ak = 4^ (k/3) if k = 0, 3, 6...
and ak equals 0 otherwise.
i dont get how they came up for the value of ak

i might be missing something silly ...
any clues?
thanks!

2. Apr 5, 2006

### dextercioby

Well, the way you asked it

$$S=\sum_{n=0}^{\infty} \left(4z^{3}\right)^{n}$$

exists iff $|4z^{3}|<1$.

Daniel.

3. Apr 5, 2006

### Galileo

They are comparing it to the general form of a power series:
$$\sum_{n=0}^\infty a_n z^n=a_0+a_1z+a_2z^2+...$$

You have
$$\sum_{n=0}^\infty 4^nz^{3n}=1+4z^3+4^2z^6+...$$

Now what is the coefficient of $$z^k$$ in the given power series? (It's clearly zero if k is not a multiple of 3).

4. Apr 5, 2006

### sweetvirgogirl

since you seem to have a good understanding on power series and such ...
could you also tell me how they expand rational functions?
like i know 1/(1-x) = 1 +x +x^2/2! ....
but how did they come up with that?

and any other interesting rules about power series expansion of rational functions?

5. Apr 5, 2006

### roger

you meant 1+x+x^2...

by binomial theorem ?

6. Apr 5, 2006

### sweetvirgogirl

oops yeah ... there is no division by n factorial involved ....

man i forgot all that power series and stuff ....

thanks!

7. Apr 6, 2006