Question

Find the values of x for which the series converges (write the answer in interval form)

Sum of ((-6)^n)(x^n) starting from n=1 to infinity is (-1/6,1/6), but now how do I find the sum of the series for these values of x?

Answer 1

do you mean it converges when x is between (-1/6, 1/6) ?

Answer 2

Yes! That's what I meant.

Answer 3

\[\sum_{n=1}^{\infty} a^n = \frac{a}{1-a}\]

Answer 4

Would I do it for both -1/16 and 1/16?

Answer 5

ofcourse the sum doesn't exist at -1/6 and 1/6

Answer 6

how did u get 16 in the denominator ?

Answer 7

right sum does not converge at those endpoints

Answer 8

I definitely meant 1/6 as opposed to 1/16

Answer 9

Okay, fine.
so we're done with the interval part of the question, rgiht ?

Answer 10

Yes! Sorry about that.

Answer 11

i see you're trying to evaluate the sum

Answer 12

Yes, I was really confused because it asked to find the sum of the series of the values of x within the interval. Would I just pick one value?

Answer 13

i think they want you evaluate the limit of sum at boundaries

Answer 14

no i think they want sum expressed in terms of x, given domain of x as (-1/6 , 1/6)

\[\sum_{n=1}^{\infty} (-6x)^n = \frac{-6x}{1-(-6x)}\]

Answer 15

\(\large \lim\limits_{x\to \frac{1}{6}^{-}} \sum\limits_{n=1}^{\infty}(-6x)^n =-1/2 \)

but the sum at other boundary doesn't exits

Answer 16

so they may be just wanting you to express the sum as function as x as @dumbcow says

Answer 17

They did want that! Can someone explain how the 6x came to be?

Answer 18

\[\large a^m \cdot b^n = (ab)^n\]

Answer 19

\[\large (-6)^n \cdot x^n = (-6x)^n\]

Answer 20

Of course, thank you again!