Find the radius of convergence $\sum_{n=1}^{\infty} (-2)^{n}\frac{x^{3n+1}}{n+1}$

Question

Find the radius of convergence $$\sum_{n=1}^{\infty} (-2)^{n}\frac{x^{3n+1}}{n+1}$$

Normally, if I have to find the radius of convergence, I'll try to transform it to power series $\sum_{n=1}^{\infty}a_nx^{n}$ and calculate limit of $|\frac{a_{n+1}}{a_n}|$

For example, $\sum_{n=1}^{\infty} (-2)^{n}\frac{x^{3n}}{n+1}$, I will let $t=x^{3}$.

But in this problem, because the index number is 3n+1, I can't do the same. So I stuck. Hope everyone helps me. Thank you so much

I suppose people are kind of not addressing OP's actual question. The OP knows about using ratio test to find the radius of convergence but is confused whether or not they can apply it in this specific scenario as they don't have a series of the form $ \sum_{n=1}^{∞} a_n x^n$ but rather they have a series of the form $\sum_{n=1}^{∞} a_n x^{3n+1}$ instead and they can't seem to reduce it to the other form to apply the ratio test. For instance, what happens when you substitute $x^{3n+1} =t^n$ so now you have the series of the form $\sum_{n=1}^∞ a_n t^n$? — William, Jun 10 '21 at 07:20
@William yeahhh, the thing you said is my problem. It really makes me confuse to solve the question :< — Jay, Jun 10 '21 at 13:21

score 2 · Answer 1 · answered Jun 10 '21 at 06:39

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Note that, if $x\ne0$,$$\left|\frac{(-2)^{n+1}\frac{x^{3(n+1)+1}}{n+2}}{(-2)^n\frac{x^{3n+1}}{n+1}}\right|=2\frac{n+1}{n+2}|x|^3$$and that therefore$$\lim_{n\to\infty}\left|\frac{(-2)^{n+1}\frac{x^{3(n+1)+1}}{n+2}}{(-2)^n\frac{x^{3n+1}}{n+1}}\right|=2|x|^3.$$So, it follows from the ratio test that the radius of convergence is $\dfrac1{\sqrt[3]2}$.

answered Jun 10 '21 at 06:39

José Carlos Santos

427,504

Hmmm. I thought if using ratio test to find the radius of convergence, we must have the series form $\sum_{n=1}^{∞} a_n x^n$ and calculus $|\frac{a_{n+1}}{a_n}|$, not $|\frac{a_{n+1}x^{n+1}}{a_nx^{n}}|$ ? Can you explain a bit more for me ? – Jay Jun 10 '21 at 13:27
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You are talking about power series of the form $\sum_{n=0}^\infty a_nx^n$ where every $a_n$ is different from $0$. Your series is$$0+0\times x+0\times x^2+0\times x^3-x^4+0\times x^5+0\times x^6+\frac43x^7+\cdots,$$and so you cannot apply it here. What I did was simply to apply the ratio test for numerical series. – José Carlos Santos Jun 10 '21 at 13:46
If you use ratio test, so you must check at the point lim = 1 ( it mean $x=\dfrac1{\sqrt[3]2}$ ) right ? Cause if limit = 1, we can't conclude anything – Jay Jun 10 '21 at 13:55
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The limit is $2|x|^3$. So, if $|x|<\frac1{\sqrt[3]2}$ the series converges and if $|x|>\frac1{\sqrt[3]2}$, the series diverges. This is enough to deduce that the radius of convergence is $\frac1{\sqrt[3]2}$. – José Carlos Santos Jun 10 '21 at 14:14
okayy, I understand, thank for your help ! Have a nice day – Jay Jun 10 '21 at 15:42
If my answer was useful, perhaps that you could mark it as the accepted one. – José Carlos Santos Jun 10 '21 at 15:54
I'm sorry, I could not find anything about the ratio test for numerical series on that link. It only speaks about $\left|\dfrac{a_{n+1}}{a_n}\right|$ and not anything about the case that you use in your answer which is, $\left|\dfrac{a_{n+1}x^{\text{something}+1}}{a_nx^{\text{something}}}\right|$ – William Jun 10 '21 at 16:12
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@William I applied the ratio test to the series $\sum_{n=1}^\infty a_n$ with $a_n=(-2)^n\frac{x^{3n+1}}{n+1}$. – José Carlos Santos Jun 10 '21 at 16:15
Well that is actually what OP was asking— The radius of convergence of a power series of the form $\sum a_n (x-a)^n$ is given by $\dfrac{1}{R}= \lim\limits_{n \to ∞} \left|\dfrac{a_{n+1}}{a_n} \right|$ hence in this case, if you take $a_n= (-2)^n \dfrac{x^{3n+1}}{n+1}$ then your series is $\sum a_n$ which is no longer a power series. – William Jun 10 '21 at 16:23
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@William And I proved that the radius of convergence is $\frac1{\sqrt[3]2}$. If you think that my proved is wrong, then tell me exactly where did I make a mistake. – José Carlos Santos Jun 10 '21 at 16:33
Oh no, don't misunderstand me. I'm not calling you wrong, i wouldn't dare. I was infact asking "how" is using that limit justified in this case, as the ratio test only gives radius of convergence of power series, (at least that's what it says in that link, I think), no? But it is no longer a power series when you choose $a_n$ like that. – William Jun 10 '21 at 16:39
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@William No. The ratio test is for numerical series. Often, it can be used for the computation of radius of convergence of power series, but it's much more general than that. – José Carlos Santos Jun 10 '21 at 16:43

score 0 · Answer 2 · answered Jun 10 '21 at 13:36

Let $\sum a_n x^n$ denote any power series (at the origin). It is wise when asking a question here to take a moment to type the statement of the theorem you propose to use. I suspect you were trying to apply this "ratio test":

Theorem $1.\quad$ If $r^{-1}\,=\,\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|$, then the radius of convergence is $r$.

It is not clear how to use Theorem 1, since you have

$$\sum a_n x^n\,=\,a_4 x_4\,+\,0 x_5\,+\,0 x_6\,+\,a_7 x_7\,+\,\dots,$$

with infinitely many $a_j = 0.$

Now let $\sum c_n$ denote any series. We have another "ratio test" in this more general context:

Theorem $2.\quad$ Assume that $\lim_{n\to\infty}\frac{c_{n+1}}{c_{n}}\,=\,\rho$, and that $|\rho|<1.$ then $\sum c_n$ converges absolutely.

In the answer provided already, it is essentially this theorem that was applied, although Santos took absolute value from the start, which is no big deal.

By basic properties of power series (and of geometric series), we can see that the radius of convergence is at least $x=\xi=2^{-1/3}.$ We can show that the radius is exactly this (in agreement with the conclusion of Santos) if we show divergence of your series for either of $x=\pm \xi.$

You can check that $(-2)^n\,=\,(-\xi)^{-3n}.$ Then you will find that we get a harmonic series upon the choice $x=-\xi.$

https://en.wikipedia.org/wiki/Power_series#Radius_of_convergence — 311411, Jun 10 '21 at 13:38
You're welcome. Be careful when saying "The Blah-Blah Rule" or the "So-and-so Theorem". People may have various different ideas of what it could be! For example, what is "The Cauchy Theorem"? (https://en.wikipedia.org/wiki/Cauchy_theorem) — 311411, Jun 10 '21 at 14:04

Find the radius of convergence $\sum_{n=1}^{\infty} (-2)^{n}\frac{x^{3n+1}}{n+1}$

2 Answers2