Differentiation involving sigma notation

Question

I am having trouble understanding the following relationship in one of my assigned problems:

$$\dfrac{d}{dx}\sum_{n = 0}^\infty \dfrac{x^n}{(n + 1)!} = \sum_{n = 1}^\infty \dfrac{nx^{n-1}}{(n + 1)!}$$

Why does the lower bound for the sigma sum change? Should it not be the case that everything pertaining to $x$ stays pertained to $x$ only, not affecting the sigma sum?

Any help would be greatly appreciated!

Edit: Upon second glance, I noticed that the term is a constant when $n = 0$, rendering the differentiated term $0$ as well. Is this why the case when $n = 0$ is omitted?

Answer 1

If $n=0$ then the term $\frac{nx^{n-1}}{(n+1)!}$ already vanishes so you don't need to include it.

So you have $$\frac{d}{dx}\sum_{n=0}^{\infty}\frac{x^n}{(n+1)!}=\frac{d}{dx}(1+\frac{x}{2!}+\frac{x^2}{3!}+....+\frac{x^n}{(n+1)!}+....)$$ $$=\frac{1}{2!}+\frac{2x}{3!}+....+\frac{nx^{n-1}}{(n+1)!}+...=\sum_{n=0}^{\infty}\frac{nx^{n-1}}{(n+1)!}=\sum_{n=1}^{\infty}\frac{nx^{n-1}}{(n+1)!}$$

Answer 2

Suppose

$$y(x)=\sum_{n=0}^{\infty}a_nx^n=a_0 + a_1x+a_2x^2+a_3x^3+\ldots$$

We can find its derivative by differentiating term by term:

$$y'(x)=\frac{d}{dx}\sum_{n=0}^{\infty}a_nx^n=\sum_{n=1}^{\infty}na_nx^{n-1}=a_1+2a_2x+3a_3x^2+\ldots$$

where the derivative of $a_nx^n$ is $na_nx^{n-1}$. Observe that the start of the summation changed from $n=0$ to $n=1$, since the constant term $a_0$ has zero as its derivative.

In your case, you can let $$a_n=\frac{1}{(n+1)!}.$$