Find the radius and the interval of convergence of the series $\sum_{n=1}^{\infty}\frac{n^3}{(n^2+1)^2}\left(\frac{4x-3}{x}\right)^n$.

Question

I was asked to find the radius and the interval of convergence of the series $\sum_{n=1}^{\infty}\frac{n^3}{(n^2+1)^2}\left(\frac{4x-3}{x}\right)^n \:,x\neq 0 \: (1)$.

This is my 1st try:

Let $X=\frac{4x-3}{x}=4-\frac{3}{x}$, then the given series $(1)$ becomes $\sum_{n=1}^{\infty}\frac{n^3}{(n^2+1)^2}X^n\: (2)$. We have

$\rho=\underset{n\rightarrow\infty}{\lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\rightarrow\infty}{\lim}\left(\frac{n+1}{n}\right)^3\left(\frac{n^2+1}{(n+1)^2+1}\right)^2|X|=|X|,$ so we get convergence if $\rho<1\Rightarrow |X|<1\Rightarrow -1<X<1$, and the radius of convergence of the series $(2)$ is $R=1$.

Now I check the endpoints. At $X=1$, the series $(2)$ becomes $\sum_{n=1}^{\infty}\frac{n^3}{(n^2+1)^2}$, this series is divergent (by using the limit comparison test with the series $\sum_{n=1}^{\infty}\frac{1}{n}$). At $X=-1$, the series $(2)$ becomes $\sum_{n=1}^{\infty}(-1)^n\frac{n^3}{(n^2+1)^2}$, this series is convergent (by using Leibniz criterion).

Thus, the radius of convergence of the series $(2)$ is $R=1$ and the interval of convergence of the series $(2)$ is $[-1;1)$.

But $X=\frac{4x-3}{x}=4-\frac{3}{x}$ and $-1\leq X<1$, so we have $-1\leq 4-\frac{3}{x}<1 \Rightarrow \frac{3}{5}\leq x<1$.

I get stuck here. Is it reasonable to say that the interval convergence of the series $(1)$ is $[\frac{3}{5};1)$? And how about the radius of convergence of the series $(1)$? I don't think $R=1$ is reasonable! So it's very nice to see your suggestions to solve this problem! Thanks.

Answer 1

Denote \begin{equation}a_{n}=\frac{n^{3}}{(n^{2} + 1)^{2}}\end{equation} and compute the upper limit of it under a square, when n goes to infinity, then divide 1 by that limit, you'll get the radius of convergence. You can find more here https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem

Answer 2

Ultimately, I would've solved this problem in the same manner, so I'll just comment on the ending tidbits:

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Is it reasonable to say that the interval convergence of the series $(1)$ is $[\frac{3}{5};1)$?

Yes. More generally, it can be shown that if $\sum_{n=1}^\infty a_n x^n$ has interval of convergence $|x|<R$, then $\sum_{n=1}^\infty a_n (f(x))^n$ has interval of convergence $|f(x)|<R$. This is an idea you exploited here.

(This is ignoring that the original series is not a power series, and hence need not have something that cleanly lines up with notions of intervals or radii of convergence.)

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And how about the radius of convergence of the series $(1)$? I don't think $R=1$ is reasonable!

Also true, this would not be a reasonable claim, since the interval has length $2/5$.

But that means that the radius of convergence is simply half of that, at $1/5$, with center at the midpoint $4/5$.

However, this is strictly in the sense of the real-number context. For power series with a radius of convergence $R$, it actually will converge for all complex numbers $z$ with $|z|<R$, not merely just the real numbers. This renders its region of convergence an open disk (plus perhaps some of the boundary). However, that is not the case for your series since it's not a power series, and in the complex numbers it will not converge on a disk but instead on an oval-shaped domain (as noted by Torsten Schoeneberg below), so the notion of a "radius of convergence" does not cleanly translate here.

Hence, to speak of a "radius of convergence" for this series only applies in a limiting context at best, and is thus a bit of an abuse of terminology.