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Is there anyone can help me recognising what's the original function of this Maclaurin series expansion below? Cheers!

$\sum\limits_{n=0}^\infty (-1)^{n+1}(n+1)\cfrac{x^{2n+1}}{(2n+2)!}$

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WolframAlpha is your friend for that: try $f$ defined by $f(x)=\frac{-\sin x}{2}$. Or you could recognize something looking like $\sin$ and go with it: $$\begin{align} \sin x &= \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}=-\sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^{2n+1}}{(2n+1)!}=-\sum_{n=0}^{\infty} (-1)^{n+1} (2n+2)\frac{x^{2n+1}}{(2n+2)!}\\&=-2 {\color{red} {\sum_{n=0}^{\infty} (-1)^{n+1} (n+1)\frac{x^{2n+1}}{(2n+2)!}}} \end{align}$$

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    $\begingroup$ Thanks so much for your prompt and correct answer! Now I wonder myself why I didn't do that before! Haha. Also thanks for pointing me out about WolframAlpha. I never used that before, but I think the time is coming for me to use it! :-) Cheers mate! $\endgroup$ – asn32 Nov 14 '16 at 22:50
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$$\begin{align}\sin(x)&=\sum_{n\ge0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}\\&=\sum_{n\ge0}(-1)^n(2n+2)\frac{x^{2n+1}}{(2n+2)(2n+1)!}\\&=-2\sum_{n\ge0}(-1)^{n+1}(n+1)\frac{x^{2n+1}}{(2n+2)!}\end{align}$$

$$\sum_{n\ge0}(-1)^{n+1}(n+1)\frac{x^{2n+1}}{(2n+2)!}=\frac{-\sin(x)}2$$

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  • $\begingroup$ @ClementC. Yeah, thanks for the catch. :D Silly me. $\endgroup$ – Simply Beautiful Art Nov 14 '16 at 21:52
  • $\begingroup$ Thanks for your answer! :-) $\endgroup$ – asn32 Nov 14 '16 at 22:50
  • $\begingroup$ @AngghaNugraha You're welcome. $\overbrace{\left(\ddot{\stackrel{\quad>}\smile}\right)}_{\begin{align}\hline\qquad\end{align}}$ $\endgroup$ – Simply Beautiful Art Nov 15 '16 at 0:06
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$$\cos x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}=\sum _{n=-1}^{\infty }{\frac {(-1)^{n+1}}{(2n+2)!}}x^{2n+2}=1+\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}}{(2n+2)!}}x^{2n+2}$$

differentiate it $$-\sin x=\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}}{(2n+2)!}}(2n+2)x^{2n+1}$$ divide by $2$ $$\frac{-\sin x}{2}=\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}}{(2n+2)!}}(n+1)x^{2n+1}$$

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