Two Taylor Series for $-\sin(x)$ are Equal But One Doesn't Work?

Question

By taking the derivative of the Taylor series of $\cos(x)$, I got two different Taylor series for $-\sin(x)$ (one starts at $n=0$, and one at $n=1$).

$$\sum_{n=1}^\infty \frac{(-1)^n\cdot x^{2n-1}}{(2n-1)!}$$ $$\sum_{n=0}^\infty \frac{(-1)^n\cdot x^{2n-1}}{(2n-1)!}$$ Graphically, these are the same, but the second series doesn't make sense to me. If you plug in $n=0$, you have to divide by $(-1)!$, which is undefined.

Can we still consider the second Taylor Series to be correct?

Answer 1

I guess you did something like this: $$\cos(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}$$ And differentiated it: $$-\sin(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n-1)!}x^{2n-1}$$ You can spot the mistake if you write out the first few terms: $$\frac{x^0}{0!}-\frac{x^2}{2!}-...$$ And you differentiated it like this: $$0 \frac{x^{-1}}{0!}-2\frac{x}{2!}+...$$ And here is the mistake: You took $\frac{0}{0!}=\frac{1}{(-1)!}$. Of course, $(-1)!$ is not defined, but because of these cases, we can take $\frac{1}{n!}=0$ for negative $n$, just like having $0^0=1$.

Answer 2

The derivative of $x^{2n}$ is $2n x^{2n-1}$.

That means that for $n=0$ it is null, and you shall cancel it (sum starting from $n=1$) before simplifying with the factorial at denominator.