# What is the radius of convergence of $\sum _{n\geq 2} n (n-1)z^{n+1}$?

So far I have used the radius of convergence test but I am not sure if it is correct:

$$\lim_{n \rightarrow \infty} \frac{\left | a_{n+1} \right | }{\left | a_{n} \right | }$$

$$\lim_{n \rightarrow \infty} \frac{{{(n+1)(n+1-1)}} z ^{n+1+1}}{(n)(n-1)z ^{n+1}}$$

$$\lim_{n \rightarrow \infty} \frac{(n+1)(z)}{(n-1)} < 1$$ i.e. when this is less than 1, the series is convergent.

Which means that $z$ has to be less than $\frac{n-1}{n+1}$

And thus the radius of convergence is $\frac{n-1}{n+1}$.

• What is $\lim_{n \to \infty} \frac{n+1}{n-1}$? That's all you need, I think. – John Lou Jul 11 '17 at 0:28
• $1/z$ has to be less than $\lim_{n\to \infty}\frac{n+1}{n-1}$ – Jonathan Davidson Jul 11 '17 at 0:31

## 2 Answers

The radius of convergence of a series is a number. Note that $$\lim_{n\to\infty}\frac{n-1}{n+1}|z| =|z|\lim_{n\to\infty}\frac{n-1}{n+1} =|z|.$$ Hence the series converges absolutely if $|z|<1$, and diverges if $|z|>1$ (and also diverges on the boundary of the disk). The radius of convergence is $1$.

• I think you mean $\displaystyle\lim_{n\to\infty}\frac{n+1}{n-1}|z|$. – lhf Jul 11 '17 at 1:28

Start with the Ratio Test:$$\lim_{n \rightarrow \infty} \frac{\left | a_{n+1} \right | }{\left | a_{n} \right | }.$$

So you get:

$$\lim_{n \rightarrow \infty} \frac{{{(n+1)(n+1-1)}} z ^{(n+1)+1}}{(n)(n-1)z ^{n+1}}$$ Which is simplified to: $$\lim_{n \rightarrow \infty} \frac{{{(n+1)(n)}} z ^{(n+2)}}{(n)(n-1)z ^{n+1}}$$

Cancel out terms, and you'll get:

$\frac{x^2}{x}$ $\lim_{n \rightarrow \infty}$ $\frac{n+1}{n-1}$

The limit of $\frac{n+1}{n-1}$ = 1, so |$x$| < 1. So the radius of convergence is 1.