Question

Find the radius of convergence and interval of convergence of the series \[\sum_{n=1}^{\infty}2^{n}(x-3)^{n}/\sqrt{n+3}\]

Answer 1

radius is 1/2

interval is [2.5,3.5)

Answer 2

how do u get the ratio test for sqrt n+3

Answer 3

I didn't use a ratio test

Answer 4

howd you do it?

Answer 5

could you do the problem if it was just \[\sum_{n=1}^{\infty}2^{n}(x-3)^{n}\]

Answer 6

yah i just dont get the sqrt

Answer 7

the radius is going to be the same

the interval will be different though

Answer 8

the \[1/\sqrt{n+3}\] really only affects the endpoints of the interval of convergence

Answer 9

note that

\[2^{n}(x-3)^{n}/\sqrt{n+3}<2^{n}(x-3)^{n}\]

so if the sum of the RRS converges on some interval then so will the LHS (plus maybe a little more)

Answer 10

if you really want to use the ratio test note that

\[\lim_{n\to\infty}\frac{\sqrt{n+3}}{\sqrt{n+1+3}}=\lim_{n\to\infty}\sqrt{\frac{n+3}{n+4}}=\sqrt{\lim_{n\to\infty}\frac{n+3}{n+4}}=\sqrt{1}=1\]

Answer 11

To answer this you will need the ratio test and apply the limit when n tends to infinity. Therefore using the definition we have:\[\left|[ 2^{n+1}(x-3)^{n+1}/\sqrt{n+1+3}]/[2^{n}(x-3)^{n}/\sqrt{n+3}] \right|\] which can be simplify as\[[2^{n+1}(x-3)^{n+1}/\sqrt{n+4}]*[\sqrt{n+3}/2^{n+1}(x-3)^{n+1}]\] which simplifies to\[\left| 2(x-3)\sqrt{n+3/n+4} \right|\] If i tend this limit to infinity we have\[\lim_{n \rightarrow \infty}\sqrt{n+4/n+3}=\sqrt{\lim_{n \rightarrow \infty}(1+4/n)/(1+3/n)}=\sqrt{1/1}=1\] which we then have that \[2\left| x-1\right|<1\] by definition of the ratio test, which the gives that \[\left| x-1 \right|<1/2\rightarrow1/2<x <2/3\]. Now to test if we should include the endpoints just evaluate for x=1/2 and x=2/3 and check if the series converges. Peace, Love and Happiness from Puerto Rico

Answer 12

Sorry a lot of typos, is late and tired its suppose to say \[2\left| x-3 \right|<1\rightarrow 2.5<x <3.5\]