# sum of the complex series

• Jun 4th 2013, 11:39 AM
alteraus
sum of the complex series
Hello everybody, I am wondering how can I find the sum of the series $\displaystyle \sum_{n=1}^{\infty}\frac{ni^n2^n}{(z+i)^{n+1}}$ ? I found out that the series converges for |z+i|>2, but I am not able to find the sum of the series , thank you very much
• Jun 4th 2013, 12:25 PM
HallsofIvy
Re: sum of the complex series
If it weren't for that "n" at the beginning, that would be a geometric series. But that n makes me think of a derivative. If we set $\displaystyle f(z)= -(2i)^n (z+ i)^{-n}$ then $\displaystyle f'(z)= n(2i)^n (z+ i)^{-n- 1}= \frac{n 2^n i^n}{(z+ i)^{n+1}}$. Does that give you any ideas?
• Jun 5th 2013, 12:43 AM
alteraus
Re: sum of the complex series
So you're saying that I could at first find the sum
$\displaystyle f(z)=\sum_{n=1}^{\infty}\Big(\frac{2i}{z+i}\Big)^n$ and I get the function f(z)
and just count the derivative of f'(z) ?
this series looks not as difficult as the first one, but I still can't find the sum ..
would you please give me another hint?
• Jun 5th 2013, 01:34 AM
Prove It
Re: sum of the complex series
Quote:

Originally Posted by alteraus
So you're saying that I could at first find the sum
$\displaystyle f(z)=\sum_{n=1}^{\infty}\Big(\frac{2i}{z+i}\Big)^n$ and I get the function f(z)
and just count the derivative of f'(z) ?
this series looks not as difficult as the first one, but I still can't find the sum ..
would you please give me another hint?

It's a geometric series...
• Jun 5th 2013, 02:05 AM
alteraus
Re: sum of the complex series
okey,
can you please explain how does the geometric series work for complex numbers?
I dealt only with real geometric series and the sum is then easy
is there any sum formula for complex terms?
in my series number $\displaystyle z \in \mathbb{C}$

thank you
• Jun 9th 2013, 07:24 AM
alteraus
Re: sum of the complex series
no more suggestions?
• Jun 9th 2013, 07:49 AM
Prove It
Re: sum of the complex series
It works exactly the same way as a real geometric series.
• Jun 9th 2013, 08:04 AM
alteraus
Re: sum of the complex series
thank you very much, I found the result which looks to be right
• Jun 9th 2013, 07:31 PM
Soroban
Re: sum of the complex series
Hello, alteraus!

I think I found the sum.
But check my work . . . please!

Quote:

$\displaystyle \sum_{n=1}^{\infty}\frac{n(2i)^n}{(z+i)^{n+1}}$

I found out that the series converges for $\displaystyle |z+i|>2$,
but I am not able to find the sum of the series.

$\displaystyle \begin{array}{cccccc} \text{We are given:} & S &=& \dfrac{1\cdot(2i)}{(z+i)^2} + \dfrac{2\cdot(2i)^2}{(z+1)^3} + \dfrac{3\cdot (2i)^3}{(z+i)^4} + \cdots \\ \text{Multiply by }\frac{2i}{z+i}\!: & \dfrac{2i}{z+i}S &=& \qquad\qquad\;\; \dfrac{1\cdot (2i)^2}{(z+i)^3} + \dfrac{2\cdot (2i)^3}{(z+i)^4} + \cdots \end{array}$

$\displaystyle \text{Subtract: }\:\left(1 - \frac{2i}{z+i}\right)S \;=\;\frac{2i}{(z+i)^2} + \frac{(2i)^2}{(z+i)^3} + \frac{(2i)^3}{(z+i)^4} + \cdots$

. . . . . . . . . . $\displaystyle \left(\frac{z-i}{z+i}\right)S \;=\;\frac{2i}{(z+i)^2}\underbrace{\left[1 + \frac{2i}{z+i} + \frac{(2i)^2}{(z+i)^2} + \cdots \right]}_{\text{geometric series}}$

The geometric series has first term $\displaystyle a = 1$ and common ratio $\displaystyle r = \frac{2i}{z+i}$
. . Its sum is: .$\displaystyle \frac{1}{1-\frac{2i}{z+i}} \:=\:\frac{z+i}{z-i}$

We have: .$\displaystyle \left(\frac{z-i}{z+i}\right)S \;=\;\frac{2i}{(z+i)^2}\cdot\frac{z+i}{z-i}$

. . . . . . . . . . . . . . $\displaystyle S \;=\;\frac{2i}{(z+i)^2}\cdot\frac{z+i}{z-i}\cdot\frac{z+i}{z-i}$

. . . . . . . . . . . . . . $\displaystyle S \;=\;\frac{2i}{(z-i)^2}$
• Jun 10th 2013, 12:53 AM
alteraus
Re: sum of the complex series
Thank you very much Soroban,
nice way how to avoid the term-by-term differentiation,
by which I got the same sum of the series, so it looks to be right :)