Question

using the ration test find the radiance of convergence for the power series \sum_{n=0}^{\infty} \frac{(n+5)x^n}{2^n+n}

Answer 1

\[\sum_{0}^{\infty} (n+5)x ^{n} \div 2 ^{n}+n\]

Answer 2

I'm sorry, I cannot remember how to do radius of convergence, but this may help: http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx

Answer 3

so i use the ration test and end up with \[x/2 \lim_{n \rightarrow \infty} n^2+6n/n^2+6n+5\]

Answer 4

\[\sum_{n=0}^\infty \frac{(n+5)x^n}{2^n+n}~~?\]

Answer 5

Oh I didn't even see the Tex code up there...

Answer 6

where did the x/2 come from on that? The tex code will help us knderstand yourwriting

Answer 7

i figured out its 2. if you get an answer k any finite number, time |x-a| the you put 1/k.
Where k=1/2 i just forgot to multiply by the reciprocal

Answer 8

Did you use the ratio test? I don't see where you may have

Answer 9

and i didnt realize that with x^n+1/x^n=1 not x

Answer 10

ok, you have me flummoxed. I do not understand what you did @SithsAndGiggles it's yours

Answer 11

Ratio test gives
\[\begin{align*}\lim_{n\to\infty}\left|\frac{(n+6)x^{n+1}}{2^{n+1}+n+1}\times\frac{2^n+n}{(n+5)x^n}\right|&=|x|\lim_{n\to\infty}\left|\frac{2^n+n}{2^{n+1}+n+1}\right|\\\\
&=|x|\lim_{n\to\infty}\left|\frac{1+n2^{-n}}{2+(n+1)2^{-n}}\right|\\\\
&=\frac{|x|}{2}\end{align*}\]
Between the last two steps, I think I can assume you can tell that
\[\lim_{n\to\infty}\frac{n}{2^n}=0\]

Answer 12

wrong

Answer 13

Sure about that? I have WA backing me up: http://www.wolframalpha.com/input/?i=Limit%5B%281%2Bn2%5E%28-n%29%29%2F%282%2Bn2%5E%28-n%29%2B2%5E%28-n%29%29%2Cn-%3EInfinity%5D

Answer 14

And me

Answer 15

you made an error in your first step. You dropped your x

Answer 16

\(\frac{x^{n+1}}{x^n} \not = 1\)

Answer 17

Well maybe I should clarify. I didn't state that the radius is 2. I just showed that, in order for the series to converge, the limit must be less than 1.

So we have \(\dfrac{|x|}{2}<1\), which gives \(|x|<2\) and indeed a radius of 2.

Answer 18

Not wrong, just not finished :P

Answer 19

then i guess its cause of the two leading n^2 numbeers where the value of the limit is the ratio of the leading coefficients of the same powers. then 1/k which is the value 1/2

Answer 20

Alright well we all seem to agree the radius is 2, right?

Answer 21

yea

Answer 22

Wonderful!

Answer 23

lol yea, siths, Just multiple ways of going about it, but @pmkat14 , you should check your work because on an exam you will get the problem wrong

Answer 24

even though you arrived at a correct answer