Question

Power Series Convergence II

Answer 1

This is the last one I promise @ganeshie8

Answer 2

Well, as I've seen by you @ganeshie8
\[\Large a_n=\frac{1}{25^{n}}\]

\[\Large \frac{1}{25}\left| x \right| \le 1\]

\[\Large \left| x \right| \le 25\]

Raidus of convergence is just 25 then

Answer 3

With that in mind:
\[\Large \frac{ 1 }{ 25} |x^2| \le 1 \]

Answer 4

\[\Large |x^2| \le 25\]

Idk how to move on forward

Answer 5

that looks good!

Answer 6

Well, I put it in my graphing calc and it shows me that the graphs of abs(x^2) and 25 intersect at +/- 5 So I would think the radius is 5

Answer 7

-5<x<5

Answer 8

Yep!

Answer 9

|x^2| < 25

is same as

x^2 < 25

because a square always produces a nonnegative number

Answer 10

Were you waiting on me to answer it myself? Lol, I guess anyway, you've taught me a lot. Now time to study for Tuesday! (AP Calc BC)

Answer 11

I can see you're fully ready :) good luck!