# Convergence radius of particular series

For the following question:

I used the ratio test and factored out the (x+2), and estimated: $$\lim_{n \to \infty}\frac{(4+(-1)^{n+1})}{(4+(-1)^n)}=\frac{3}{5}$$

Now from what I understand, I should multiply $$\frac{3}{5}(x+2)$$ and find -11/3 < x < -1/3, and radius=$$\frac{5}{3}$$.

However, the correct answer seems to be to leave out the $$\frac{3}{5}$$ altogether and find that -3 < x < -1, meaning the radius=1.

What am I not understanding?

• I think I got the wrong limit :\ Is it actually infinity? – blizz Apr 6 at 1:28
• The limit is not well defined. The fraction in the limit flicks between $\frac35$ and $\frac53$. So you take the "worst" one, which is the largest one. – John Doe Apr 6 at 1:28
• @JohnDoe thank you, I figured that out too late, but even if I take the larger limit, I still end up with the wrong radius and interval... – blizz Apr 6 at 1:31

The limit$$\lim_{n \to \infty}\frac{(4+(-1)^{n+1})}{(4+(-1)^n)}$$ is not defined (not necessarily infinite). You could take the largest value of the two that the fraction changes between, but there is a less restrictive way to go, and in fact the sum converges for more values of $$x$$ than you'd get if you used $$\frac53$$.

Write $$\sum_n(4+(-1)^n)(x+2)^{n-1}=4\sum_n(x+2)^{n-1}+\sum_n(-1)^n(x+2)^{n-1}$$If you do ratio test on these two, you find that the radius of convergence is $$1$$ for both power series. This gives you your answer.

Alternatively (as Robert Shore pointed out), to determine convergence at the boundaries, it may be clearer to split the series up as $$3\sum_n(x+2)^{2n}+5\sum_n(x+2)^{2n-1}$$and see the radii of convergence here (also $$1$$). Since these are both positive, and the series both diverge at $$x=-3,-1$$, so does their sum. In the previous way of splitting them up, we had some positive and some negative terms, and it may have looked like the divergences at the boundaries could cancel. This clarifies that they indeed don't.

• I think there's something wrong here. The second sum is an alternating series that converges whenever $x \gt -1$. It seems to me the right way to break up the sum is the odd terms and the even terms. – Robert Shore Apr 6 at 1:54
• @RobertShore Are you sure? The alternating series test requires each successive term to decrease monotonically to $0$ for convergence, which would only happen if $|x+2|<1$, no? – John Doe Apr 6 at 1:58
• Right. I got it backwards. But I'm still concerned with convergence, particularly at the boundaries (the question requests the interval of convergence as well as the radius of convergence). It still seems cleaner to me to break up the sum into even and odd terms, because then you know you're dealing exclusively with positive terms. – Robert Shore Apr 6 at 2:00
• @RobertShore Yes, I suppose that would be much clearer. But the result would be the same, so what do you mean about the boundaries? – John Doe Apr 6 at 2:10
• Maybe it's because I'm tired but it's not immediately obvious to me from the fact that neither of your series converges at the boundaries ($-3$ and $-1$) that their sum also fails to converge. It is obvious to me that if neither of two strictly positive series converges, their sum also must diverge. – Robert Shore Apr 6 at 14:37

Note that in such a case as yours the ratio test is inconclusive and one can only say that the radius of convergence lies somewhere between $$\frac{3}{5}$$ and $$\frac{5}{3}$$.

The root test is here more powerful:

• $$r = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}}$$

With $$a_n = 4+(-1)^n$$ you get

$$1 \leq \sqrt[n]{4+(-1)^n} \leq \sqrt[n]{5} \stackrel{n \to \infty}{\longrightarrow} 1$$

Hence,

$$r = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{4+(-1)^n}} = \frac{1}{\lim_{n \to \infty} \sqrt[n]{4+(-1)^n}} = \frac{1}{1} = 1$$