For the Power series

$$\sum_{n=1}^{\infty} 3^{n}(x+4)^{n}/(n^{1/2})$$

By applying the ratio test I get

3|x+4|

so I determined a = -4
how do I find the radius (R)
some how my prof got 1/3 I do not understand how he came upon such a solution

Question

For the Power series

$$\sum_{n=1}^{\infty} 3^{n}(x+4)^{n}/(n^{1/2})$$

By applying the ratio test I get

3|x+4|

so I determined a = -4
how do I find the radius (R)
some how my prof got 1/3 I do not understand how he came upon such a solution

turingtest · Answer

the series will converge if $$3|x+4|<1\implies|x+4|=\frac13$$so R=1/3 is the radius of convergence

australopithecus · Answer

isn't the radius the distance of

-4 to the end points of the interval?

turingtest · Answer

no, I think you are thinking of the interval of validity, which is$$-\frac13<x+4<\frac13$$$$-\frac{13}3<x<-\frac{11}3$$

australopithecus · Answer

man I'm confused lol

australopithecus · Answer

is there a general method to compute the radius?

turingtest · Answer

look at what the ratio test is looking for...

australopithecus · Answer

it is looking for the area of convergence, for which values of x make the function convergent (although the end points need to be checked)

australopithecus · Answer

sorry by function I mean series

turingtest · Answer

when$$\lim_{n\to\infty}\left|{\{a_{n+1}\}\over\{a_n\}}\right|<1$$we get convergence, so we take that limit, and if the limit exists we call it $L$
we then have something that looks like$$L|f(x)|<1$$the radius of convergence is the range in which $|f(x)|$ coonverges, which is$$|f(x)|<\frac1L$$

turingtest · Answer

...to find the interval of validity you would solve the above for x

australopithecus · Answer

ok thanks :) I think I get it now

turingtest · Answer

just in case you want further info http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

australopithecus · Answer

thanks for the link I will definitely check it out

turingtest · Answer

welcome!