Mike S.

asked • 08/13/17# Find the interval of convergence of the power series

∞

n=0

## 2 Answers By Expert Tutors

Dom V. answered • 08/23/17

Cornell Engineering grad specializing in advanced math subjects

**(something in terms of x)<1**. Then we can identify which values of x actually satisfy the inequality.

^{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).

- First a bit of algebraic rearrangement

^{n})(x

^{n}) 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.

^{(n+1)})*x

^{(n+1)}]/[2(n!/3

^{n})x

^{n}] | <1

- Simplify everything inside the limit (2s cancel; (n+1)!/n! = n+1; 3
^{n}/3^{(n+1)}=1/3; x^{(n+1)}/x^{n}=x)

- 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)

- 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).

**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.

Andy C. answered • 08/13/17

Math/Physics Tutor

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

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

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

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

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

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

------------ 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

Mike S.

08/13/17

Andy C.

08/14/17

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Andy C.

08/14/17