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Determine the sum of the series

$$\sum_{n=2}^\infty \frac{(-1)^nn}{(2n)!} x^{2n}$$

I realize that there is a sum by comparison for $\cos x$ which is defined by

$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n} = \cos x$$

However, how would I go about converting it to this form as the answer I get does not conform with the posted solution

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4  
What happens when you differentiate? –  Potato Dec 2 '12 at 3:33

1 Answer 1

up vote 2 down vote accepted

Start with $$\cos x=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}$$ and differentiate with respect to $x$ to get

$$\begin{align*} -\sin x&=\sum_{n\ge 0}\frac{(-1)^n(2n)}{(2n)!}x^{2n-1}\\ &=2\sum_{n\ge 0}\frac{(-1)^nn}{(2n)!}x^{2n-1}\;. \end{align*}$$

Now multiply by $x$ and make any necessary adjustments in the range of indices.

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