1. ## series

Hey All,

Can someone check if I have done this right. Thank you

find the range of values for x for which:
____
s = 9^nx^{2n} / n^3
__n=1

Un + 1 = 9^n.9.x^{2n}.x^2 / n^3 + 3n^2 + 3n + 1

(Un+1)/(Un)= 9^n.9.x^{2n}.x^2 / n^3 + 3n^2 + 3n + 1 . n^3 / 9^n + x^{2n}

simplifying

9x^2.n^3 / n^3 + 3n^2 + 3n + 1

dividing n^3

9x^2 / 1 + 3/n + 3n^2 + 1/n^3

n→1

9x^2 / 8

2. Originally Posted by dadon
Hey All,

Can someone check if I have done this right. Thank you

find the range of values for x for which:
____
s = 9^nx^{2n} / n^3
__n=1

Un + 1 = 9^n.9.x^{2n}.x^2 / n^3 + 3n^2 + 3n + 1

(Un+1)/(Un)= 9^n.9.x^{2n}.x^2 / n^3 + 3n^2 + 3n + 1 . n^3 / 9^n + x^{2n}

simplifying

9x^2.n^3 / n^3 + 3n^2 + 3n + 1

dividing n^3

9x^2 / 1 + 3/n + 3n^2 + 1/n^3

n→1

9x^2 / 8
Not sure how you ended up with 9x^2 / 8, but i suppose you want the values of x for which S converges.

Here,

3. how do you know it is less than 1?

thanks

4. Originally Posted by dadon
how do you know it is less than 1?

thanks
Because the theorem says that the ratio limit must be strictly less than 1 for the series to converge.