Expressing $\sum_{n=0}^{\infty} (\frac{z}{1+z})^n$ as a power series

Question

I'm basically given the following task:

Express the series:

$$\sum_{n=0}^{\infty} \left(\frac{z}{1+z}\right)^n$$

as a power series for all $z \in \mathbb{C}$ where the series converges.

So, I know I need to get the above series in the form $$\sum_{n=0}^{\infty} a_n(z)^n$$ where the $a_n$ are constants.

I've tried my best to algebraically manipulate the summand to no success! I know that $$\frac{z}{1+z} = \frac{1}{1+\frac{1}{z}} = \frac{1}{1-(-\frac{1}{z})} = \sum_{j=0}^{\infty} (-1)^j\left(\frac{1}{z}\right)^j$$

but plugging any of these into the summand of the given series only seems to make things worse!

I've tried to expand out the original series:

$$\sum_{n=o}^{\infty} \left(\frac{z}{1+z}\right)^n = 1 + \frac{z}{1+z} + \frac{z^2}{(1+z)^2} + \ldots $$

but I don't see a pattern.

Any suggestions?

Answer 1

Proceeding naively, in $f(z) =\sum_{n=0}^{\infty} \left(\frac{z}{1+z}\right)^n $, we must have $\left|\frac{z}{1+z}\right| < 1$.

Then $f(z) =\sum_{n=0}^{\infty} \left(\frac{z}{1+z}\right)^n =\dfrac{1}{1-\frac{z}{1+z}} =\dfrac{1+z}{1+z-z} =1+z $.

Check:

If $z=1$ then $\frac{z}{1+z} =\dfrac12 $ so $f(z) =\sum_{n=0}^{\infty} (1/2)^n =\dfrac1{1-1/2} =2 $ and $1+1 = 2$, so, OK.

Answer 2

You may do this: $$\sum_{n=0}^{\infty} \left(\frac{z}{1+z}\right)^n = \frac{1}{1-\frac{z}{1+z}} = \frac{1+z}{1+z-z} = 1+z$$

Answer 3

Let $$f(z)= \sum _0^\infty (\frac {z}{1+z})^n = 1+ (\frac {z}{1+z}) + (\frac {z}{1+z})^2+ ...$$

We have $$f(z) = f(0) + f'(0)z + f''(0)/2 z^2 +.....$$

It is easy to see that $$f(0)=1, f'(0)=1, f''(0)=0, f'''(0) =0,...$$

Thus $$f(z) = 1+z$$