# Power series

Hi
I"m having truoble with fnding the power series of the following:

1+(x^3)/3+(x^6)/18+(x^9)/162+...

Can anyone give me a hand?

Also, any hints in finding a power series?

Thanks !

Hurkyl
Staff Emeritus
Gold Member
That IS a power series...

quasar987
Homework Helper
Gold Member
Hahaha... or did you mean, you have trouble putting it into compact $\sum$ notation?

Tide
Homework Helper
I am guessing that overseastar is looking for a closed form of the sum.

Looks like:

$$\sum_{n=0}^{a}\frac{x^{3n}}{3^n3!}$$

Joffe said:
Looks like:

$$\sum_{n=0}^{a}\frac{x^{3n}}{3^n3!}$$

Why do you use the factorial when 3 ! = 6 which is constant ?

Joffe said:
Looks like:

$$\sum_{n=0}^{a}\frac{x^{3n}}{3^n3!}$$
That fails to work when the denominator is 3.

I think he means $$\sum_{n=0}^{\infty}\frac{x^{3n}}{3^nn!}$$

Thankyou Moo Of Doom, that is what I meant to write.

HallsofIvy
Homework Helper
And $$\sum_{n=0}^{\infty}\frac{x^{3n}}{3^nn!}$$
is equal to
$$\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^3}{3}\right)^n$$
Anyone recognize THAT??

No I don't, please enlighten me.

$$\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^3}{3}\right)^n=\exp{\left(\frac{x^3}{3}\right)}$$

oh wow, yes, that's what i meant, sorry

And thank you for all your help !