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 $\frac{n-1}{n+1}$.

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

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 $1$.

  • $\begingroup$ I think you mean $\displaystyle\lim_{n\to\infty}\frac{n+1}{n-1}|z|$. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.