Question

series to function and find the interval of convergence

Answer 1

\[\sum_{1}^{\infty} ((x^{2k})/4k)\]

Answer 2

\[\lim_{k \to \infty}\frac{4k}{4k+4}=1\] so you need
\[|x^2|<1\]

Answer 3

can you tell you how you got to the 4k/4k+4

Answer 4

He applied the ratio test:\[ \lim_{k \rightarrow \infty} \frac{x^{2(k+1)}}{4k+1} \frac{4k}{x^{2k}}\]

Answer 5

Remember that by the Ratio test, the series will converge if that limit above is < 1, so we get \(|x^2| < 1\)

Answer 6

Typo, it should be 4(k+1) not 4k + 1

Answer 7

BTW is you do ration dont you get xk/k+1 ???

Answer 8

? I am sorry, I don't understand it. You should get:\[\lim_{k \rightarrow \infty} |x^2|\frac{4k}{4k+1} = |x^2| < 1\]

Answer 9

Gah, again that typo, it's 4(k+1) or 4k + 4

Answer 10

how do you get your fraction to be top bottom instead of side by side ex. 1/3

Answer 11

so we can simplify k/k+1 right?

Answer 12

Yeah, the limit is 1, so we are left with |x^2| < 1. And write \frac{numerator}{denominator} on the equation button and it will write like that.

Answer 13

\frac{1}{2}

Answer 14

Click the equation button right below here and type it there.

Answer 15

\(\frac{1}{2}\) like this

Answer 16

aswsome thanks, but i think we loose the x^2 when we cross mult. the \[\frac{a_{k+1}}{a_{k}}\]

Answer 17

No, one of them is x^2(k+1) = x^2*x^2k the other is 2k, we are left with x^2.

Answer 18

Something like \(\LARGE \frac{x^2 x^{2k}}{x^{2k}}\)

Answer 19

your are right... i think its nap time my brain needs the recharge. ive been going at this for far to many hours. thanks for your help

Answer 20

No problem and have a nice rest.