# I need help with finding the radius of convergence of a complex series

1. May 3, 2012

### taxmccall13

Radius of convergence of the series n^2(x^n)/(3n!) I am stumped

the question is: find the radius and interval of convergence of the following series {sum_(n=1)^(Infinity)}((n^2)(x^n))/(3*6*9***3n)

I'm assuming that equal to ((n^2)(x^n))/(3n)!

then lim_(n->infinity) of (((n+1)^2(x^(n+1))/(3n+3)!)*((3n)!/((n^2)(x^n))

= lim_(n->infinity) of ((n+1)^2(x))/((n^2)(3n+3)(3n+2)(3n+1))
=x*lim_(n->infinity) of ((n+1)^2)/(27n^5+54n^4+33n^3+6n^2)
lim_(n->infinity) of (n^2)/(n^5) --> 0
=x*0 = 0<1

so now what? i don't know how to find the radius of convergence in this situation

does it help to know that the {sum_(n=1)^(Infinity)}(x^n)/n! converges to e^x?

my book scarcely mentions factorials in series, is there a good place to review them?

Last edited: May 3, 2012
2. May 3, 2012

### taxmccall13

nevermind the denominator is n!*3, silly me

Last edited: May 3, 2012
3. May 3, 2012

### Dick

Think again. Isn't 3^n*n! more like it?

4. May 3, 2012

### taxmccall13

wouldn't 3^n*n! give
N Value
1 3
2 18
3 162
giving (3*18*163***3^3n*3n)

are those the values you should get, my book lacks anything explaining what 3*6*9***3n would even mean.

I thought it meant 3(n!) because the first two are right
N Value
1 3
2 6 but then
3 18
3n 9n
giving (3*6*18***9n)

Is there are pattern for these, I tried to find them in my book and on Wikipedia but i don't know what this expression is called

Last edited: May 3, 2012
5. May 3, 2012

### Dick

E.g. 3*6*9=3*1*3*2*3*3=(3*3*3)*(1*2*3)=3^3*3!. You seem to be confusing the individual terms with the whole product.

6. May 3, 2012

### taxmccall13

Wow, it makes so much more sense now, thank you
I just didn't know what it all meant. you are my hero for the day

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