Given series $\sum_{n=0}^{\infty}a_nz^n$ where $a_n=\begin{cases}\dfrac{1}{3^n} & \text{when $n$ is even} \\ \dfrac{1}{5^n} & \text{when $n$ is odd} \end{cases}$
Find the radius of convergence.
My work:
So I take the even and odd parts separately, and calculate the radius of convergence individually using the ratio test. Then I got radius of convergence $3$ for the even terms and $5$ for the odd terms. Now I don't know which will qualify as the radius of convergence of the whole series. The answer says $3$ is the r.o.c. So is it the smaller one which always qualifies? Help me to understand this concept. Thanks.
Best Answer
The radius of convergence is the smaller radius -- 3. The evens only converge if $|z| < 3$ and the odds only converge if $|z| < 5$. Thus, both the evens and odds converge only if $|z| < 3$, and you need both of them to converge for the combined series to converge.
Think about it this way, if you let $z = 4$ (or any number between 3 and 5), then the odd terms will be okay and converge to some finite value but the even terms will blow up (i.e., diverge). The sum of something that blows up and a finite value is something that blows up.
If $|z| < 3$, then both the even and odd series converge, and the combined series converges to the sum of whatever the even and odd series converge to.
If $|z| > 5$, then both the even and odd series diverge and thus the combined series diverges.