# Determine (or evaluate) the sum of the series $\sum_{k=1}^\infty \frac{x^{2k}}{(2k+1)k!}$.

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.

$$x\sum_{k=1}^\infty \frac{x^{2k}}{(2k+1)k!}=\sum_{k=1}^\infty\int\dfrac{(x^2)^k}{k!}$$
$$\sum_{k=1}^\infty\dfrac{(x^2)^k}{k!}=e^{x^2}-1$$
• 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
• Thank you. But, $\int (e^{x^2}-1) dx$ isn't know. – Mark Jan 16 '16 at 10:48