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How to evaluate series $$\sum_{n=0}^\infty\frac{5n+1}{(2n+1)!}$$ I tried to split the summation...but I failed. Please help

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Hint: $$5n+1 = \frac{5}{2}(2n+1) - \frac{3}{2}$$

Do you know $\sum_{n=0}^\infty \frac{1}{(2n+1)!}$ and $\sum_{n=0}^\infty \frac{1}{(2n)!}$ ?

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  • $\begingroup$ Thank you. We can group the terms accordingly to get the answer $\frac{e}{2}+\frac{2}{e}$ $\endgroup$ – user538954 May 21 at 16:44
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$$\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$$ $$\frac{1}{x}\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n+1)!}$$ let $x\rightarrow \sqrt{x}$ $$\frac{1}{\sqrt{x}}\sinh \sqrt{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{(2n+1)!}$$

let $x\rightarrow x^5$ $$\frac{1}{\sqrt{x^5}}\sinh \sqrt{x^5}=\sum_{n=0}^{\infty}\frac{x^{5n}}{(2n+1)!}$$ multiply by $x$ $$\frac{x}{\sqrt{x^5}}\sinh \sqrt{x^5}=\sum_{n=0}^{\infty}\frac{x^{5n+1}}{(2n+1)!}$$ $$(\frac{x}{\sqrt{x^5}}\sinh \sqrt{x^5})'=\sum_{n=0}^{\infty}\frac{(5n+1)x^{5n}}{(2n+1)!}$$ then let $x=1$ to get the sum

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