# How to find the radius of convergence for such series

Suppose we have a series $$\sum^{\infty}_{n=1} \frac{(z+i)^{3n}}{n^2}$$

Since $z$ is a complex term, can we use root test in this case?

I tried both root test and ratio test, both give me the radius of convergence $\rho = 1$, by having $a_n = \frac{1}{n^2}$. Can someone tell me if I'm on the right track or not? Many thanks

• you are on the right track – Doug M May 10 '16 at 22:31
• What also concerns me is that, in order to apply the root test, $n$ in $a_n$ must match the power of $z^n$, in this case it doesn't? but why it produces the same result as ratio test does? Or I misunderstood something? – Gvxfjørt May 10 '16 at 22:38
• Not really on the precisely right track - you should be concerned about what you say you're concerned about. You got (un?)lucky, getting the right answer just because $1^{1/3}=1$. In fact $a_n=(n/3)^{-2}$ if $n$ is divisible by $3$, $a_n=0$ if $n$ is not divisible by $3$. If you apply the root test to that sequence you should get the right answer by a correct method. (You need the version of the root test with $\limsup$ or $\liminf$, whatever; a version with $\lim$ will not be applicable.) – David C. Ullrich May 10 '16 at 22:51
• For this series, the Ratio Test (not a consequence of it, but the actual test where you divide the $(n+1)st$ summand by the $n$th summand and take the limit of the modulus of that ratio) works perfectly well. – Greg Martin May 10 '16 at 23:13

## 1 Answer

Letting $t = (z+ \Bbb i) ^3$ transforms the series into $\sum _{n \ge 1} \frac 1 {n^2} t^n$. The root test tells us that the radius of convergence is $\lim _{n \to \infty} \sqrt[n] {\frac 1 {n^2}} = 1$, so for $|t| < 1$ the series converges and for $|t| > 1$ it diverges.

If $t=1$ the series converges and if $|t| = 1$ and $t \ne 1$ then we may use Dirichlet's test: the sequence $(\frac 1 {n^2}) _{n \ge 1}$ decreases to $0$ while

$$\left| \sum _{n =1} ^N t^n \right| = \left| \frac {1 - t^{N+1}} {1-t} -1 \right| \le \frac {1 + |t^{N+1}|} {|1-t|} + 1 \le \frac 2 {|1-t|} + 1 ,$$

a bound independent of $N$, so Dirichlet's test tells us that for such $t$ the series converges.

We have discovered that the series in $t$ converges for $|t| \le 1$, in other words for $|z+ \Bbb i|^3 \le 1$, which means $|z+ \Bbb i| \le 1$, which is the disc of center $\Bbb i$ and radius $1$.

• This is a clear and thorough answer +1 – Andres Mejia Nov 4 '16 at 19:26