# Taylor series of $\frac{\sin x}{x}$

I wanted to find the Taylor series of the function $$\frac{\sin x }{x}$$ and found on wolfram alpha that it is equal to, in series representation, $$\sum^{\infty}_{n=-1} \frac{x^n((-i)^n+i^n)}{2(1+n)!}.$$

Now this might be a stupid question but why is it not just the case of applying $\frac{1}{x}$ to the Taylor series of $\sin x$? Then finding it to be $$\sum^{\infty}_{k=0} \frac{1}{x} \frac{(-1)^kx^{1+2k}}{(1+2k)!} = \sum^{\infty}_{k=0} \frac{(-1)^kx^{2k}}{(1+2k)!}.$$

• It's the same thing. – Jack Feb 12 '18 at 15:31
• You've taken $\frac{1}{n}$ inside the infinite sum, but this is only valid if the resulting sum converges. – SecretlyAnEconomist Feb 12 '18 at 15:36
• Do note that if you look at the "Series expansion" part of WA for $\sin x$ and then divide it by $x$ you get exactly the expansion for "Series expansion" of $\frac {\sin x} x$ – Gaurang Tandon Feb 12 '18 at 15:39

These two are in fact equal. Notice that $$\frac{(-i)^n+(i)^n}{2}$$ is $$0$$ when $$n$$ is odd and $$\pm1$$ when $$n$$ is even. So you can substitute $$n=2k$$.
The CAS (computer algebra system) sometimes look to work thing out in a no optimal way, of course the results should be the same (yes, you can divide the sin(x) McLaurin series by x and get the power series for $sin(x)/x$, with $x\neq0$), so I wonder that's what happens.