Question

what is the interval of convergence for the following expression?

Answer 1

\[\sum_{n=0}^{\infty}\left(\begin{matrix}(x ^{3}-2)^{2n} \\ 4^{n}\end{matrix}\right)\]

Answer 2

is that spose to be a fraction? or combinatorics?

Answer 3

fraction

Answer 4

...yeah, there is no fraction option in the eq editor is there

Answer 5

i noticed. its quite annoying

Answer 6

frac{}{} for further uses if you need it ;)

Answer 7

lets flip the rule over (its reciprocal); and subtract 1 form all the ns

Answer 8

\[\frac{(x^3-2)^{2n}}{4^n}*\frac{4^{n-1}}{(x^3-2)^{2(n-1)}}\]

and simplify

Answer 9

\[\lim_{n\to\ inf}\frac{(x^3-2)^2}{4}\]

pull out all the nonN parts

\[\left| \frac{(x^3-2)^2}{4}\right|\lim_{n\to\ inf}1\]

Answer 10

set this up in the convergence formualtion

\[|\frac{(x^3-2)^2}{4}|<1\]

and solve for x

Answer 11

i never would have thought of using the reciprocal... thanks

Answer 12

your welcome