Can You Help Me Find a Power Series?

[overseastar]

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]

That IS a power series...

 

[quasar987]

Hahaha... or did you mean, you have trouble putting it into compact $\sum$ notation?

 

[Tide]

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

 

[Joffe]

Looks like:

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

 

[roger]

Looks like:

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

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

 

[amcavoy]

Looks like:

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

That fails to work when the denominator is 3.

 

[Moo Of Doom]

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

 

[Joffe]

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

 

[HallsofIvy]

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??

 

[Joffe]

No I don't, please enlighten me.

 

[amcavoy]

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

 

[overseastar]

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

 

[overseastar]

And thank you for all your help !