# What is the radius of convergence of $i + \dfrac{2}{1-i}\displaystyle\sum_{j=1}^{\infty} \biggl(\dfrac{z-i}{1-i}\biggr)^{j}$?

When I apply the ratio test, I get:

$$\displaystyle\lim_{j \rightarrow \infty}\biggl| \dfrac{2(z-i)^{j+1}(1-i)^{j+1}}{(1-i)^{j+2}(z-i)^{j}}\biggr|$$=$$\displaystyle\lim_{j \rightarrow \infty} \biggl|\frac{2(z-i)}{(1-i)}\biggr|$$. Thus giving me that that the radius of convergence is

$$|z-i|<\dfrac{\sqrt{2}}{2}$$? I'm not sure if I'm doing the process correctly?

• The $2$ should also appear in the denominator. – user647486 Apr 14 at 18:29

Why not use substitution? With $$\;u=\dfrac{z-1}{1-i}$$, you obtain the series in $$u$$: $$1+\frac 2{1-i}\sum_{j=1}^\infty u^j,$$ which converges if and only $$|u|<1$$, i.e. $$\;|z-i|<|1-i|=\sqrt 2$$.

• whenever I'm finding the radius of convergence, do I have to take into account whatever is multiplied with the summation? so, $\dfrac{2}{1-i}$ in this case? – K.M Apr 14 at 18:44
• No, unless it is a series. – Bernard Apr 14 at 18:45

I don't know where does that $$2$$ come from. Your approach is correct (the root test would work too) and the radius of convergence is $$\sqrt2\left(=\lvert 1-i\rvert\right)$$.

• I took the $2$ outside the summation and included it with $\dfrac{(z-i)^{j}}{(1-i)^{j}}$ – K.M Apr 14 at 18:38
• But if you put it in the numerator, you have to put it in the denominator too. – José Carlos Santos Apr 14 at 18:55

Alternatively:

The series is well known to have unit radius of convergence for the argument $$\dfrac{x-i}{1-i}$$.

Then

$$\left|\dfrac{x-i}{1-i}\right|<1\iff|x-i|<\sqrt2.$$