Calculating the radius of convergence for $\sum _{n=1}^{\infty}\frac{\left(\sqrt{ n^2+n}-\sqrt{n^2+1}\right)^n}{n^2}z^n$

Question

Calculate the radius of convergence for $$\sum _{n=1}^{\infty}\frac{\left(\sqrt{ n^2+n}-\sqrt{n^2+1}\right)^n}{n^2}z^n.$$

Do I have to do a index shift so the sum starts at $0$? I tried it like here (without index shift) Finding the_radius of convergence, and my solution for the denominator is $+\infty$ so in the end my radius of convergence should be $0$.

Is that correct? Thanks!

Answer 1

No, you are not correct. In order to find the radius of convergence you are supposed to evaluate the reciprocal of the limit of $|a_n|^{1/n}$ (the so-called root-test) where $a_n$ is the coefficient of $z^n$. Now note that as $n\to+\infty$, $$\sqrt{ n^2+n}-\sqrt{n^2+1}=\frac{(n^2+n)-(n^2+1)}{\sqrt{ n^2+n}+\sqrt{n^2+1}}=\frac{1-\frac{1}{n}}{\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{1}{n^2}}}\to \frac{1}{2}.$$ Can you take it from here and find the correct radius of convergence?

Answer 2

Note that $$\Big(\frac{1}{n^2}(\sqrt{ n^2+n}-\sqrt{n^2+1})^n\Big)^{1/n}=\frac{\sqrt{ n^2+n}-\sqrt{n^2+1}}{(n^{1/n})^2}$$ Now $\lim_{n\to\infty}n^{1/n}=1$ and $\lim_{n\to\infty}(\sqrt{ n^2+n}-\sqrt{n^2+1})=\frac12$. Hence $$\lim_{n\to\infty}\Big(\frac{1}{n^2}(\sqrt{ n^2+n}-\sqrt{n^2+1})^n\Big)^{1/n}=\frac12$$ You need to use the Root Test now.