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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

  • 4
    What happens when you differentiate?2012-12-02

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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.