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?

• There's something called a constant of integration. – Michael Hardy Apr 30 '14 at 17:00
• When differentiating the series you changed the lower index of the summation from $i=0$ to $i=1$ to $i=0$, and back to $i=1$. In all cases it should remain as $i=0$, and there is no issue here. – Perry Elliott-Iverson Apr 30 '14 at 17:07
• I agree with Peter Iverson. You probably thought that you should move the sumation for one index forward from the 3rd to the 4rd row while doing the derivation. What you did not have to do. – WalyKu Apr 30 '14 at 17:11

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 $$\frac{d}{dx}\sum_{n=0}^\infty \frac{x^{n+1}}{n+1} = \sum_{n=0}^\infty x^{n}$$