I want to know the interval of the convergence of the following power series.

Question

I have a power series which is $$\sum_{n=0}^\infty (2+(-1)^n)^n x^n$$ I tried to apply ratio test for it and the solution i have; they applied Cauchy test and the radius of convergence determined was $1/3$ and i could not understand how? Please explain.

Answer 1

Hint: You need to use $\displaystyle \limsup_{n \to \infty}$ not $\displaystyle \lim_{n \to \infty}$ for the root test.

https://en.wikipedia.org/wiki/Root_test

$\displaystyle \limsup_{n \to \infty} \sqrt[n]{\lvert \left( 2 + (-1)^n \right)^n x^n \rvert} = \limsup_{n \to \infty} \lvert 2 + (-1)^n \rvert |x| = 3|x|$. So the root test says :

if $3|x|>1$ then the series diverges,

if $3|x|=1$ then the test is inconclusive,

if $3|x|<1$ then the series is convergent.

Thus, for $\displaystyle -\frac{1}{3}<x<\frac{1}{3}$ we have convergence for sure. To check the endpoints:

i) $\displaystyle x=\frac{1}{3} \implies \sum_{n=1}^{\infty} \left( 2 + (-1)^n \right)^n x^n = \sum_{n=1}^{\infty} \left( 2 + (-1)^n \right)^n \frac{1}{3^n} = \sum_{n=1}^{\infty} \left( \frac{2 + (-1)^n}{3} \right)^n$ diverges by the general ($n$th) term test since $\displaystyle \left( \frac{2 + (-1)^n}{3} \right)^n = 1$ for even $n$, $\displaystyle \lim_{n \to \infty} \left( \frac{2 + (-1)^n}{3} \right)^n \neq 0$.

ii) $\displaystyle x=-\frac{1}{3}$ case can be treated similarly to conclude the series diverges too in this case.

Hence, the interval of the convergence is $\displaystyle -\frac{1}{3}<x<\frac{1}{3}$ and the radius of the convergence is $\dfrac{1}{3}$.