Using the limsup of a power series to find the radius of convergence

Question

I have a hard time understanding why the limsup of the power series should be $3$. I tried googling and solving this problem on my own for quite some time now but I cant get it right :/

Can someone please explain this to me?

Answer 1

In these radius of convergence problems we can ignore anything that grows in "polynomial time."

That is, you can ignore this part $\frac {(2n^3-2)}{17n^2+5}$

We can do this because $\lim_\limits{n\to\infty} \sqrt [n]{\frac {(2n^3-2)}{17n^2+5}} = 1$

That leaves the factors $3^n(x-2)^{2n}$

We need $|\lim_\limits{n\to\infty} \sqrt [n] {3^n(x-2)^{2n}}| < 1$

$|3(x-2)^2|<1\\ |x-2|<\frac {1}{\sqrt {3}}$

If you don't like the root test, you can use the ratio test.

$\lim_\limits{n\to\infty} \left |\frac {\frac {(2(n+1)^3 - 2)3^{n+1}(x-2)^{2(n+1)}}{17(n+1)^2 + 5}}{\frac {(2n^3 - 2)3^{n}(x-2)^{2n}}{17n^2 + 5}}\right| < 1$

$\lim_\limits{n\to\infty} \left|\frac {2(n+1)^3 - 2}{2n - 2}\frac {17n^2+5}{17(n+1)^2 + 5}3(x+2)^2\right| < 1$

Answer 2

You have

$$\frac{1}{n} \ln \left( \frac{2n^3-2}{17n^2+5}\right) = \frac{1}{n} \ln (2n^3-2) - \frac{1}{n} \ln(17n^2+5)$$

tends to $0$, so $$\sqrt[n]{\frac{|2n^3-2|}{17n^2+5}} = \exp \left( \frac{1}{n} \ln \left( \frac{|2n^3-2|}{17n^2+5}\right)\right)$$

tends to $1$, and therefore

$$\sqrt[n]{\frac{|2n^3-2|3^n}{17n^2+5}} =\left(\sqrt[n]{\frac{|2n^3-2|}{17n^2+5}} \right) \times 3$$

tends to $3$.