Find the interval of convergence of the power series

Question

Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval. If the interval of convergence is an interval, enter your answer using interval notation. If the interval of convergence is a finite set, enter your answer using set notation.)

∞

∑  (2n!)(x/3)^n
n=0

Answer 1

For most interval of convergence questions, you should first attempt either a ratio test (a must if there are factorials present, but still usually the easiest to pull off in any scenario) or a root test.  

 

RATIO TEST: lim_(n->inf) [a_(n+1)/a_n] = R. Converges when R<1, diverges when R>1, inconclusive when R=1.

ROOT TEST: lim_(n->inf) [(a_n)^(1/n)] = R. Same conclusions based on value of R.

 

These tests are useful because they have an explicit way of finding the "border" between converging/diverging behavior. When finding radius of convergence, we'll apply one of the tests and actually restrict it to values <1 (forcing it to converge). Because we're looking at a power series instead of a regular infinite series, the limit will evaluate to (something in terms of x)<1. Then we can identify which values of x actually satisfy the inequality.

 

Power series are written as a sum over a_n*(x-c)ⁿ. The a_n are coefficients, and c is the "center point" of the power series (plugging in x=c eliminates all terms except for a_0). When you apply the ratio/root tests you want to arrange things from (something in terms of x)<1 into |x-c|<r. Whatever value r turns out to be is your radius of convergence. The absolute value signs are there on purpose. Applying the ratio/root test to the absolute value of the series will guarantee absolute convergence, and you won't have to worry about tricky algebraic exceptions that come up when dealing with alternating series (because |(-1)ⁿ|=1 no matter what).

 

 

Your example has factorials present, so we default to the ratio test.

• First a bit of algebraic rearrangement

(2n!)(x/3)^n  = 2(n!/3ⁿ)(xⁿ)             I want the (x-c)ⁿ by itself. c=0 here.

• Apply root test on the absolute value of the power series; set the limit to values <1.

lim_(n->inf): | [2((n+1)!/3⁽ⁿ⁺¹⁾)*x⁽ⁿ⁺¹⁾]/[2(n!/3ⁿ)xⁿ] | <1

• Simplify everything inside the limit (2s cancel; (n+1)!/n! = n+1; 3ⁿ/3⁽ⁿ⁺¹⁾=1/3; x⁽ⁿ⁺¹⁾/xⁿ=x)

lim_(n->inf): | [(n+1)/3]*x | <1

• Evaluate the limit and put into |x-c|<r form (I'm going to treat infinity like a number just so you can see what happens)

|∞*x|<1  or  |x|<1/∞  or |x|<0

• Your radius of convergence is therefore r=0.
• Your interval of convergence extends from c-r < x < c+r. Since r=0, your power series only converges at the single point x=c (x=0). 

If we had gotten any other nonzero value for r, we would have defined an open interval where the power series converges absolutely. The next step would be to test the two endpoints of the interval to see if the entire interval of convergence was truly open, closed, or half-open. Simply plug each endpoint value into x in the original power series formula, and then apply one of the standard convergence tests to the resulting infinite series.

Answer 2

(2(n+1))!(x/3)^(n+1)

------------------- =

(2n)!(x/3)^(n)


(2n +2)! (x/3)^(n+1)
--------------------- =
(2n)! (x/3)^n


(2n+2)(2n+1)(x/3)

 

DIVERGES

------------ UPDATE: 8/14/17 at 03:18

I am doubling down on my answer.
The series diverges.

https://www.symbolab.com/solver/series-calculator
That link is to a series convergence calculator, which upon
request will show all of the steps.
It did the exact same ratio test I did, and got exactly the
same result. The limit blows up and the series diverges.

The radius of convergence is zero. {0}
that's it