We have to find the radius of convergence of the given power series,
$$\sum_{n=0}^{\infty} (-1)^n \frac{n^{2n}}{(4n+1)^n}(x+2)^{n^2}$$
I think the only way to solve this might be the root test but all I'm getting is that
$\ \lim_{n\to \infty} \frac{n^2|x+2|^n}{4n+1} \le 1 $ for convergence.
and from there we can say that if |x+2|>1 then the series is definitely divergent, but if it is less than 1 then we have the 0. $\infty $ form which I'm not able to solve.
Could someone please confirm if what I've done is right and also give me a hint as to how to proceed ?
Thanks
Best Answer
You are right about the series diverging for $|x+2|>1$. To show it converges when $|x+2|<1$, intuitively you have $\frac{n^2}{4n+1}$ which goes to $\infty$ at $\mathcal{O}(n)$ and $|x-2|^n$ which goes to zero exponentially, so it dominates and the product goes to zero. You can prove this rigorously with L'Hopital for instance by flipping appropriate terms: $$\lim_{n\to\infty}\frac{n^2|x-2|^n}{4n+1}=\lim_{n\to\infty}\frac{n^2}{(4n+1)\left(\frac{1}{|x-2|}\right)^n}$$ Notice now the numerator and denominator both go to infinity. So take the derivative of both to get an equivalent limit: $$\lim_{n\to\infty}\frac{2n}{\left((4n+1)\log\left(\frac{1}{|x-2|}\right)+4\right)\left(\frac{1}{|x-2|}\right)^n}$$ Both still go to infinity so derive again: $$\lim_{n\to\infty}\frac{2}{\left((4n+1)\log\left(\frac{1}{|x-2|}\right)+8\right)\log\left(\frac{1}{|x-2|}\right)\left(\frac{1}{|x-2|}\right)^n}$$ Now the numerator is constant but the denominator goes to infinity, so the limit is zero and we have rigorously verified that the series converges when $|x-2|<1$