Question

For what values of x does the series \(\sum_{n=1}^\infty \frac{(4x-5)^{2n+1}}{n^{\frac{3}{2}}}\) converge absolutely? Conditionally?

Answer 1

is the denominator
\[\sqrt{n^3}\]?

Answer 2

Yep.

Answer 3

let me try again

Answer 4

For the interval of convergence I got 1 < x < 3/2, if that helps. I'm also not sure how to analyze the convergence/divergence at the endpoints.

Answer 5

i am being foolish
you need ratio test for this one, should get the limit is 1 so you have to solve
\[|4x-5|<1\]

Answer 6

oh ok you are right

Answer 7

so lets replace x by 1 and see what happens

Answer 8

you get
\[\sum\frac{(-1)^{2n+1}}{\sqrt{n^3}}\]which certainly converges because it is alternating and the terms go to zero
it also converges absolutely because the exponent on the denominator is greater than one

Answer 9

replace x by 3/2 and you get
\[\sum\frac{1}{\sqrt{n^3}}\] which also converges for the same reason above
so it converges at the endpoints of the interval as well

Answer 10

How do you determine what values of x converge conditionally?

Answer 11

ok first off if it converges absolutely it converges conditionally

Answer 12

you already have the interval of convergence, you computed it as
\[1<x<1.5\]
now at the left hand endpoint you get an alternating sequence, so it is possible that it converges "conditionally" meaning that the absolute value does not converge.
but in this case the absolute value does converge because the exponent in the denominator is greater than one