Proving Convergence of sum[n=0 to inf] (z+2)^(n-1)/((n+1)^3 * 4^n) for |z+2|<=4

[jaci55555]

Homework Statement

Prove that the series sum[n=0 to inf] (z+2)^(n-1)/((n+1)^3 * 4^n)
converges for |z+2| <=4

Homework Equations

The Attempt at a Solution

sum[n=0 to inf] (z+2)^(n-1)/((n+1)^3 * 4^n) <= sum[n=0 to inf] |(z+2)^(n-1)/((n+1)^3 * 4^n)|
<= sum[n=0 to inf] |(z+2)|^(n-1)/(|(n+1)|^3 * 4^n)
<= sum[n=0 to inf] (4)^(n-1)/(|(n+1)|^3 * 4^n)
<= sum[n=0 to inf] 1/((n+1)^3 * 4)

 

[eczeno]

hello,

your reasoning looks sound, although it is a bit hard to read, so I am not totally sure, but you should state why you know that last series converges. a slight problem is that your proof doesn't show that these are the only values of z for which the series converges. of course, the problem does not ask for that, so maybe it is not an issue. the ratio test would just spit out the radius of convergence so you wouldn't have to use it in your proof. that might be a 'nicer' way to prove the convergence.

cheers

 

[jaci55555]

Thank you for answering, I'm still figuring LaTeX out :)
so should i use what i did, and then the ratio test? I did this and got
lim (0-> inf) ((n+1)^3/(n+2)^3), are there any steps i should put after this, or is it fine that i can see that this is <1 ?
Thanks again for you help!

 

[eczeno]

ok, so I see how you got that limit, but there should be a factor of \frac{\left|{x-2}\right|}{4} multiplied by the limit. it is this product that must be less than one, which is how you will determine the radius of convergence. the limit you have is actually not less than one, it is exactly equal to one.

 

[jaci55555]

ok, so I see how you got that limit, but there should be a factor of \frac{\left|{x-2}\right|}{4} multiplied by the limit. it is this product that must be less than one, which is how you will determine the radius of convergence. the limit you have is actually not less than one, it is exactly equal to one.

So then lim(n->inf) (z+2)/4 <1
so that lim(n->inf) (z+2) <4 but i need to find find <=4

 

[eczeno]

you are there. the limit should not be there anymore, we took it already and it equaled one. even with it in there, though, the statement is correct (well, almost) because |z+2| just stays |z+2| no matter what n does. so the limit as n->inf of |z+2| is just |z+2|. you do need to put absolute value around the z+2 though, just putting (z+2) makes it wrong.

 

[jaci55555]

so i remove the lim and then |Z+2|<4, but i still need it to be equal to 4 as well

 

[HallsofIvy]

Look at the cases z=-6 and z= 2 (z+2= -4 and z+ 2= 4) separately.