Recall that:

The power series for #1/(1-x)# is #sum_(n=0)^N x^n#.

What you can do is take the derivative of both sides:

#d/(dx)[sum_(n=0)^N x^n] = d/(dx)[1/(1-x)] = color(green)(d/(dx)[1 + x + x^2 + ...])#

When we take the derivative, note that the first term will disappear (#d/(dx)[1] = 0#), so the first #n# shifts from #n=0# to #n=1#.

#sum_(n=1)^N nx^(n-1) = 1/(1-x)^2 = color(green)(?)#

We can shift it back to #n = 0# for conventional purposes.

How you do that is consider that you are currently starting at #n=1#. That means #n-1 = 0#. To get the exponent to become #0# using #n = 0# below #sum#, #n-1# becomes #n#.

When #n = 1# below #sum#, #nx^(n-1)# starts at #1x^0#, so when #n = 0# below #sum#, #nx^"stuff"# becomes #(n+1)x^"stuff"#.

#color(highlight)(sum_(n=0)^N (n+1)x^(n) = 1/(1-x)^2) = color(green)(?)#

So, all we need to do is do it for the explicit series to get:

#1/(1-x)^2 = d/(dx)[1 + x + x^2 + ...]#

#= color(blue)(1 + 2x + 3x^2 + 4x^3 + ...)#

The radius of convergence is based on the idea that the sum of magnitudes #|a| >= 1# will not converge if they are all positive.

#1/(1-x)^2 > 1# when #|x| >= 1#.

Therefore, the radius of converge is:

#color(blue)(x in (-1, 1))#

or you can write it as:

#color(blue)(|x| < 1)#