Determine (or evaluate) the sum of the series $$\sum_{k=1}^\infty \frac{x^{2k}}{(2k+1)k!}, \ \ \ x\in\mathbb R^{+}$$ The best that I have managed to do is $$\sum_{k=1}^\infty \frac{x^{2k}}{(2k+1)k!}<\sum_{k=1}^\infty \frac{\left(x^{2}\right)^k}{(k+1)!}=\frac{1}{x^2}\left(e^{x^2}-x^2-1\right)$$ Thanks in advance.
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HINT:
$$x\sum_{k=1}^\infty \frac{x^{2k}}{(2k+1)k!}=\sum_{k=1}^\infty\int\dfrac{(x^2)^k}{k!}$$
$$\sum_{k=1}^\infty\dfrac{(x^2)^k}{k!}=e^{x^2}-1$$
lab bhattacharjee
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Excuse me, but maybe it is: $\sum_{k=1}^\infty \frac{x^{2k}}{(2k+1)k!}=\frac{1}{x}\sum_{k=1}^\infty\int\dfrac{(x^2)^k}{k!}$. – Mark Jan 16 '16 at 10:44
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@Mark, Thanks for your observation. – lab bhattacharjee Jan 16 '16 at 10:46
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Thank you. But, $\int (e^{x^2}-1) dx$ isn't know. – Mark Jan 16 '16 at 10:48
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@Mark, See http://math.stackexchange.com/questions/270721/how-to-evaluate-the-integral-int-ex3dx – lab bhattacharjee Jan 16 '16 at 10:51