Question

\[\sum_{n=1}^{\infty} \frac{10^{n}x^{n}}{n^{3}}\]

Answer 1

I have the radius of convergence as 1 and
\[\lim_{n \rightarrow \infty} \left|x\right| \frac{10n^{3}}{(n+1)^{3}}\]

Answer 2

i still need to find the interval of convergence

Answer 3

quick question...wouldn't R always have to be <1 to be convergent?

Answer 4

try using limit test

Answer 5

isn't that what I did...I guess I didn't complete it

Answer 6

\[\lim_{n \rightarrow \infty} \left|x\right| \frac{10}{(1+1/n)^{3}}\]

Answer 7

Hmm ... that gives, radius of convergence is 1/10

Answer 8

why 1/10 and not 10

Answer 9

if x were 10 ... then just imagine what would happen to your series ... plug this into wolfram alpha and see ... the series will be divergent.

Answer 10

ok...so now the interval is (1/10,1/10]

Answer 11

im sorry

Answer 12

(-1/10,1/10]

Answer 13

haha

Answer 14

no ... that's your radius of convergence.

Answer 15

sigh...ok lets see

Answer 16

does it have to do with numerator and denominator with the n terms (n+1)^3 and n^3

Answer 17

no i'll take that back

Answer 18

ok so i test the end points of R to see if it converges

Answer 19

wait ... i'll be right back.

Answer 20

- infty to +1/10 because it doesnt matter how small n becomes

Answer 21

k

Answer 22

it seems that your series converges for
|x| < 1/10
so your interval of convergence is -1/10 < x < 1/10

Answer 23

how is that different from (-1/10,1/10]

Answer 24

or (-1/10,1/10)

Answer 25

1/10 is closed on right hand side ..

Answer 26

(-1/10,1/10)

Answer 27

let's do it from the start.