Finding the sum of the series using differentiation

Question

How can the left hand side be differentiated when it is defined only over integers and is not continuous and hence not differentiable?

Answer 1

It is not stating that the function defined over only the integers. The function itself is just a series of polynomials where $x \in \mathbb{R}$, which is differentiable. $i$ is just an integer parameter to yield your polynomial series.

$$\sum_{i=0}^n (x^i)=1+x+x^2+x^3+...+x^n$$

Since $x \in \mathbb{R}$, this can be differentiated to give:

$$1+2x+3x^2+...+nx^{n-1}$$

Which is:

$$\sum_{i=0}^n (ix^{i-1})$$

Which was the answer obtained.

Answer 2

One may recall that $$ (f_1+f_2)'=f'_1+f'_2 \tag1 $$ giving by induction $$ (f_1+f_2+\cdots+f_n)'=f'_1+f'_2+\cdots+f'_n \tag2 $$ where we assumed all $f_i$ differentiable on a given set.

Observe that you can rewrite $(2)$ as follows $$ \left(\sum_{i=1}^nf_i\right)'=\sum_{i=1}^nf'_i. \tag3 $$

Now take $f_i$ such that $f_i(x)=x^i$, then $f'_i(x)=ix^{i-1}$ and $(3)$ rewrites $$ \left(\sum_{i=1}^nx^i\right)'=\sum_{i=1}^nix^{i-1}. \tag4 $$ Hoping it helps.