Derivatives of Series and the Fundamental Theorem of Calculus (Part 1) The first part of the Fundamental Theorem of Calculus (FTC) states that:
$$\frac{d}{dx}\int f(x)\,dx=f(x)$$
meaning that the indefinite integral of a function can be reversed by its equivalent derivative. What I can't figure out is how this is applied to infinite series. For example, given an infinite series:  
$$ \sum_{n=0}^{\infty} x^n$$ 
taking the integral: 
$$\int \sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty} \int x^n = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}+C $$
Now, theoretically, taking the derivative should result in the original series. However, to take a derivative, $n_{0}$ increases by $1$ becoming $n = 1$. So we end up with:  
$$ \frac{d}{dx} \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} +C$$ 
$$= \sum_{n=1}^{\infty} \frac{d}{dx}\left[ \frac{x^{n+1}}{n+1}+C \right]$$ 
$$= \sum_{n=1}^{\infty} \frac{n+1}{1}\frac{x^{n+1-1}}{n+1}$$ 
$$= \sum_{n=1}^{\infty} x^n$$
$$\Rightarrow  \sum_{n=0}^{\infty} x^n= \sum_{n=1}^{\infty} x^n$$
However, we know this isn't the case, because by definition, 
$$ \sum_{n=0}^{\infty} x^n= x^0 +\sum_{n=1}^{\infty} x^n = 1+ \sum_{n=1}^{\infty} x^n$$ and therefore, 
$$ \sum_{n=0}^{\infty} x^n \neq \sum_{n=1}^{\infty} x^n$$
What is the logic behind doing derivatives this way, and how does it work with part one of FTC?
 A: The issue is that your statement
$$
\frac{d}{dx}\sum_{n=0}^\infty \frac{x^{n+1}}{n+1} = \sum_{n=1}^\infty \frac{d}{dx}\Big(\frac{x^{n+1}}{n+1}\Big)
$$
is not correct. Write out the first few terms of the left-hand side:
$$
\frac{d}{dx}\Big(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots \Big)
$$
whereas on the right-hand side you have
$$
\frac{d}{dx}\frac{x^{1+1}}{1+1} + \frac{d}{dx}\frac{x^{2+1}}{2+1} + \cdots
$$
which is not the same. You have incorrectly shifted summation indices in that step. The problem your statement is that 
$$\frac{d}{dx}\sum_{n=0}^\infty a_n = \sum_{n=1}^\infty \frac{d}{dx}a_n
$$
only applies when the first number in the sequence is a constant (like in $\sum_{n=0}^\infty x^{n}$), since you're actually taking the derivative of the series as it is in expanded form, not summation form. Since the derivative of a constant is $0$, taking the derivative of a series where the first number is a constant shifts the summation indices. This is why 
$$
\frac{d}{dx}\sum_{n=0}^\infty x^{n}  = \sum_{n=1}^\infty nx^{n-1}
$$
but
$$
\frac{d}{dx}\sum_{n=0}^\infty \frac{x^{n+1}}{n+1} = \sum_{n=0}^\infty x^{n}
$$
