If f(x)=∑(n=1 to ∞) (-1)^n*(x-π)^n/n^2 find f '''(x)

So I'm trying to find the 3rd derivative of f(x), these are the steps I've taken trying to differentiate it in the summation form:

f '(x)=∑(n=2 to ∞) (-1)^n*(x-π)^(n-1)/n

f ''(x)=∑(n=3 to ∞) (-1)^n*(n-1)*(x-π)^(n-2)/n

f '''(x)=∑(n=4 to ∞) (-1)^n*(n-1)*(n-2)*(x-π)^(n-3)/n

My first question is do I move up the counter by 1 every time I take the derivative? Also do I have to keep incrementing it up for the (-1)^n as well?

Secondly, when I do term by term differentiation I get the following series for the 3rd derivative:

{-2/3 + 3(x-π)/2 - 12(x-π)^2/5 + 10(x-π)^3/3 +...}

This doesn't match the series that I get from the above sum, any ideas where I'm screwing up? Thanks

you need not move up the counter for first derivative

It is actually

f '(x)=∑(n=1 to ∞) (-1)^n*(x-π)^(n-1)/n

f ''(x)=∑(n=2 to ∞) (-1)^n*(n-1)*(x-π)^(n-2)/n

f '''(x)=∑(n=3 to ∞) (-1)^n*(n-1)*(n-2)*(x-π)^(n-3)/n

now it matches with term by term differentiation

To your first question, yes and no. You start moving up by 1 after you take the first derivative because you see, the term n=1 for f'(x) still exists, it's a scalar. The reason you move it up by 1 in the seceding derivatives is because this (and those of the following derivatives) scalar becomes 0. So n=2 for f''(x) and n=3 for f'''(x)

You don't move the counter unless the first term is 0.

f '(x)=∑(n=1 to ∞) (-1)^n (x - π)^(n-1)/n, the index still starts at 1, since the first term is nonzero.

f ''(x)=∑(n=2 to ∞) (-1)^n (n-1)*(x - π)^(n-2)/n, since n-1 = 0 for n = 1.

f '''(x)=∑(n=3 to ∞) (-1)^n (n-1)(n-2) (x - π)^(n-3)/n, since n-2 = 0 for n = 2.
.......= -2/3 + 3(x - π)/2 - 12(x - π)^2/5 + ..., as required.

I hope this helps!