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.

  • $\begingroup$ Yes, you need to take the most "stringent" radius of convergence (and the same is true for the interval of convergence). Kind of consider the interval of convergence as a "domain" on which series converge. You want your series to converge for your variable. So you should choose for the ROC the value of $3$ $\endgroup$
    – imranfat
    Jul 20 '16 at 2:47
  • $\begingroup$ @imranfat can you elaborate a bit more. I didn't get the full picture. $\endgroup$ Jul 20 '16 at 2:55
  • $\begingroup$ You want to have your series to be convergent for $z$ for even terms as well as odd terms. Now I am going to be a little "weird" here, but $1/5^n$ is "stronger" convergent than $1/3^n$ Reason? There are values of $z$ that will make the situation with $1/3^n$ divergent and with $1/5^n$ not. Please don't quote me literally on the word "stronger" $\endgroup$
    – imranfat
    Jul 20 '16 at 2:58

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.

  • $\begingroup$ Now if I take radius of convergence as $3$ and $z=4$, then how the series will converge? $\endgroup$ Jul 20 '16 at 2:51
  • 1
    $\begingroup$ No, it will diverge, thanks to the even terms $\endgroup$
    – imranfat
    Jul 20 '16 at 2:55
  • $\begingroup$ @HarryPotter I edited my response, I hope it's a bit more clear $\endgroup$
    – nukeguy
    Jul 20 '16 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.