The convergence of the geometric series

I am working out the radius of convergence of the z^n series by taking the limit of the difference of partial sums.

I am really struggling to see how 1 + z + ... + z^k - z - z^2 -... - z^(k+1) = (1 - z^(k+1)) / (1 - z)

I understand the rest of the problem and did do this last year, but I can't find my notes.

I am assuming you want to find:
(1 + z + ... + z^k) - [z + z^2 + ... + z^(k + 1)].

The first series is geometric with first term 1 and common ratio z, so:
1 + z + ... + z^k = 1[1 - z^(k + 1)](1 - z)
= [1 - z^(k + 1)]/(1 - z).

The second series is geometric with first z and common ratio z, so:
z + z^2 + ... + z^(k + 1) = z[1 - z^(k + 1)]/(1 - z).

Then, the required difference is:
[1 - z^(k + 1)]/(1 - z) - z[1 - z^(k + 1)]/(1 - z)
= [1 - z^(k + 1)][1/(1 - z) - z/(1 - z)]
= [1 - z^(k + 1)](1 - z)/(1 - z)
= 1 - z^(k + 1).

I hope this helps!

For the equation you have written, it seems to equal 1-z^(k+1) on the left, and on the right you have divided it by (1-z) for no reason! It seems to work well in the case z=0 only!

Here is the basic trick about how these sums are handled:

For a partial sum S(k) = 1 + z + z^2 + ... + z^k notice zS(k) = S(k) - 1 + z^(k+1).
Then solve for S(k).
It should give you S(k) = (1 - z^(k+1))/(1-z) = (z^(k+1)-1)/(z-1).

Given that it converges, the infinite sum is a number.
This happens when |z|<1.
Call it S
Then notice S-1= zS and solve for S.
It gives S = 1/(1-z).