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I need to calculate the sum $\sum_{n=0}^{\infty}{(-1)^n}\frac{2^{2n-1}}{(2n)!}$ . It seems very "similar" to Taylor expansion of functions arcsin(x) and its derivative for x = -2.

It is known: $arcsin(x) = \sum_{n=0}^{\infty}\frac{{(2n-1)!!x^{2n+1}}}{2^{n}n!(2n+1)}$.

When derivative applied we get: $\frac{1}{\sqrt{1-x^{2}}} = \sum_{n=1}^{\infty}\frac{{(2n+1)(2n-1)!!x^{2n}}}{2^{n}n!(2n+1)} = \sum_{n=1}^{\infty}\frac{{(2n-1)!!x^{2n}}}{2^{n}n!}$.

What to do next? Am I on the wrong trace here?

Thank you all in advance!

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    $\begingroup$ $\frac{cos(2)}{2}$ $\endgroup$
    – d.k.o.
    Jun 23 '15 at 1:14
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    $\begingroup$ It's even more "similar" to, $$\cos(x)=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}$$ Divide it by $2$ and substitute $x=2$ to get your result (as shown by d.k.o) $\endgroup$ Jun 23 '15 at 1:41
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It surprises me to see that you are familiar with the rather obscure series of $\arcsin x$, and yet, at the same time, fail to recognize one of the most well-known series in math, namely that of $\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}~.~$ Also, $\sin x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}~.$ Remember that $e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}~,~$ and then recall Euler's formula $e^{ix}=\cos x+i\sin x$.

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  • $\begingroup$ @Zach466920 I'm interested in the definitions of "linear" knowledge and "thinking too linearly." Perhaps the OP has seen the TS for the cosine function and just for a moment, didn't recognize it herein. $\endgroup$
    – Mark Viola
    Jun 23 '15 at 4:29
  • $\begingroup$ @Zach466920 I didn't ask for "definitions for the words," I stated that I'm interested in your definitions of phrases (e.g., "linear knowledge"). What is "linear knowledge?" And a "moment" is not limited to an hour+, so my conjecture is viable. My comment is not intended to provoke debate or create unhealthy tension. Rather, I hope each of us can be more empathetic towards problems others have in appealing for help. While your math skills are better than some, they are dwarfed by best-in-class mathematicians. Would you enjoy having them publicly criticize you? Humility is a key here. $\endgroup$
    – Mark Viola
    Jun 23 '15 at 14:41
  • $\begingroup$ @Zach466920 No worry. And I do understand the sentient better now. Best wishes and "happy mathing!" $\endgroup$
    – Mark Viola
    Jun 23 '15 at 14:52

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