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I am given the following series and asked to solve it. $\sum\limits_{n=1}^\infty \dfrac{(-2)^n}{(2n+1)!}$ I recognize that this series is somewhat similar to the Taylor series for $sinx$ which is $\sum\limits_{n=0}^\infty (-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$.

However, I am not really able to relate these two series in order to solve them, especially since my series starts at 1 and once I rewrite the $sinx$ series to match that, I am completely lost.

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Since $\sin(x)-x =\sum\limits_{n=1}^\infty (-1)^n\dfrac{x^{2n+1}}{(2n+1)!} $, $\dfrac{\sin(x)-x}{x} =\sum\limits_{n=1}^\infty (-1)^n\dfrac{x^{2n}}{(2n+1)!} =\sum\limits_{n=1}^\infty \dfrac{(-x^2)^{n}}{(2n+1)!} $.

Therefore, putting $\sqrt{x}$ for $x$, $\dfrac{\sin(\sqrt{x})-\sqrt{x}}{\sqrt{x}} =\sum\limits_{n=1}^\infty \dfrac{(-x)^{n}}{(2n+1)!} $.

And there you are.

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    $\begingroup$ Subtract $1$ because the series actually starts at $n=1$, not $n=0$, so the first term of the expansion is ignored. $\endgroup$ – Noble Mushtak Apr 19 '16 at 23:50
  • $\begingroup$ Oh wow. Thank you so much, makes it very clear. $\endgroup$ – jessicajjensen Apr 19 '16 at 23:50
  • $\begingroup$ @ Noble Mushtak: Thanks for the correction. $\endgroup$ – marty cohen Apr 19 '16 at 23:55
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    $\begingroup$ Shouldn't it be $sin(x)-x$ instead of $sinx-1$?? Because the first term is an x. $\endgroup$ – jessicajjensen Apr 20 '16 at 0:58
  • $\begingroup$ You are right. Thanks. I'm 0 for 2 today. Fixed. $\endgroup$ – marty cohen Apr 20 '16 at 2:36

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