Answer
$\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n}}{(2n+1)!}$
Work Step by Step
We know that the Maclaurin series for $\sin x=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+..... $
Divide the above series by $x$.
$\dfrac{\sin x}{x}=\dfrac{x-\dfrac{(x)^3}{3!}+\dfrac{(x)^5}{5!}+.....}{x}=1-\dfrac{x^2}{6}+\dfrac{x^4}{120}-.........$
Thus, our required series is:
$\dfrac{\sin x}{x}=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n}}{(2n+1)!}$
