Question

Power Series

Answer 1

When calculating the interval I'm getting (-1/2) and (1/2)

Answer 2

@agent0smith

Answer 3

What I did:
\[\sum_{}^{}\frac{ x^n }{ 2^n }=\sum_{}^{}((\frac{ x }{ 2 })^n)^\frac{ 1 }{ n }=\left| x \right|\lim_{n \rightarrow \infty}\frac{ 1 }{ 2 }\rightarrow =-\frac{ 1 }{ 2 }< x< \frac{ 1 }{ 2 }\]

Answer 4

Oh I think I see my mistake.

Answer 5

What you did is about on the right track, you just wrote it a little wrongly.

Answer 6

should be \[|\frac{ x }{ 2 }| < 1\]

Answer 7

I am not quite sure you could write the following:
\(\sum_{}^{}\frac{ x^n }{ 2^n }=\sum_{}^{}((\frac{ x }{ 2 })^n)^\color{red}{\frac{ 1 }{ n }}=\left| x \right|\lim_{n \rightarrow \infty}\frac{ 1 }{ 2 }\rightarrow =-\frac{ 1 }{ 2 }< x< \frac{ 1 }{ 2 }\)
The offending part is in red, which violates equality.
Also, I think the question is understood to be "what is the interval of convergence for x", and not the complete term.

Answer 8

\[\large \sum_{}^{}\frac{ x^n }{ 2^n }=\sum_{}^{} \left( \frac{ x }{ 2 } \right)^n\]

Answer 9

I think your 1/n was a typo... just idk how you managed it :/

Answer 10

no not a typo.
I applied the root test for series

Answer 11

Then you can remove the part to the left of the first equal sign.

Answer 12

so i should have omitted the sum is what you mean?

Answer 13

You know that:
\(\color{blue}{\displaystyle \sum_{}^{}\frac{ x^n }{ 2^n }=\sum_{}^{} \left( \frac{ x }{ 2 } \right)^n}\)

and from there all you need is an elementary test for a geometric series:
Set \(\color{blue}{\displaystyle \left|\frac{x}{2}\right|<1}\), and solve for \(\color{blue}{\displaystyle x}\).

Answer 14

i think this is what's being offered.

Answer 15

Oh ok, I see. I dont really have to do root test here.

Answer 16

Yup, you don't:)

Answer 17

For geometric series, if the number in the parentheeses = 1 it diverges right?

Answer 18

Yes

Answer 19

because your terms are all the same, and essentially you will be adding \(a_0\) forever.

Answer 20

Thank you everyone

Answer 21

No problem:)

Answer 22

(by the way, your initial conclusion is incorrect.)

Answer 23

I thought you recognized it as a geometric series, hence thinking the 1/n was just a minor brain malfunction with the algebra

Answer 24

x/2 is the geometric ratio, so as long as |x/2|<1 the series will converge.

Answer 25

\(\color{blue}{\displaystyle \left|x/2\right|=1\quad \Longrightarrow \quad \left|x\right|=2}\).