Find the maximal ring where the following series converges: $$\sum_{n=1}^\infty\frac{3^n+2^n}{(z-5)^n}+\sum_{n=0}^{\infty}\frac{n^2}{20^n}(z-5)^{2n}$$

I think that taking the minimum between the two separate radiuss' will be enough. The right series is taylor series which is entire and analytic in $\mathbb C$ and $$\displaystyle{\overline{\lim}_{n\to\infty}\frac {\sqrt[n]{n^2}}{20}}=\frac 1 {20}$$ so $|z-5|<0.05$ but for the second series (the laurent series which has main singularity in z=5), I cannot find how to calculate the radius of convergence. How can I do so? (and by the way, am I correct with the taylor series?)

  • $\begingroup$ The radius of convergence is for taylor series. Do you mean region of convergence, maybe? $\endgroup$ – Adam Hughes Jul 14 '14 at 4:55
  • $\begingroup$ if region means $A(z_0,r_1,r_2)$ yes. $\endgroup$ – user65985 Jul 14 '14 at 4:57
  • $\begingroup$ Ah, then that's not bad. See my answer below. $\endgroup$ – Adam Hughes Jul 14 '14 at 5:00

For the first one, you can use comparison test* to realize this is just a question of

$$\lim\sup_{n\to\infty}\left|{3^n\over (z-5)^n}\right|^{1/n}<1\iff {3\over |z-5|}<1\iff |z-5|>3$$

Similarly the second one gives:

$$\limsup_{n\to\infty}\left|{n^2(z-5)^{2n}\over 20^n}\right|^{1/n}<1\iff |z-5|^2<20\iff |z-5|<\sqrt{20}$$

so you just get $3<|z-5|<2\sqrt 5$.

Note the two inequalities point in different directions, this is because you are getting an annulus, it's not about a minimum, because we're not talking about two Taylor series that converge on a disc with a common center, we're talking about it diverging inside one disc and outside the other.

*The comparison is that $3^n+2^n<2\cdot 3^n$ so that

$$\sum_{n=1}^\infty \left|{3^n+2^n\over (z-5)^n}\right|<2\sum_{n=1}^\infty \left|{3^n\over (z-5)^n}\right|$$

Note that it is justified because we ask about convergence at each point in the annulus, so we can assume $z$ is fixed, and each term still represents some number, just like with series without variables.

  • $\begingroup$ About the first: you compare it to $\sum\frac{3^n}{(z-5)^n}$? if so: why this test is correct in this context in the complex? about the second: why you put $(z-5)^{2n}$ inside? $\endgroup$ – user65985 Jul 14 '14 at 5:01
  • $\begingroup$ @CoarguAliquis The test for convergence requires you take the $n$th root of the $n$th term, but the term *includes* the stuff with $z$, so you put it in as well. I have already included the comparison at the bottom of my answer as you were writing this comment, so take a look and let me know if that answers your question. $\endgroup$ – Adam Hughes Jul 14 '14 at 5:03
  • $\begingroup$ I think I got it. one last question: why did we need the right inequality (i.e $\sum_{n=1}^\infty\big|\frac{3^n}{(z-5)^n}\big|<\sum\big|\frac{3^n+2^n}{(z-5)^n}\big|$)? $\endgroup$ – user65985 Jul 14 '14 at 5:06
  • $\begingroup$ You don't really, I suppose: you can ignore it. $\endgroup$ – Adam Hughes Jul 14 '14 at 5:08
  • $\begingroup$ Now I got it. Thanks again. $\endgroup$ – user65985 Jul 14 '14 at 5:08

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