Evaluate the Sum

Question

Evaluate $$\sum_{n=0}^{\infty}\frac{5n+1}{(2n+1)!}$$

I tried using $e^{x}=\sum_{k=0}^{\infty}\frac{x^k}{k!}$ But I got $$5e-8\sum_{n=0}^{\infty}\frac{1}{2n+1}$$ Hints would be appreciated!

Answer 1

$5n+1=\dfrac52\left(2n+1\right)+1-\dfrac52$

$$\implies \sum_{n=0}^{\infty}\frac{5n+1}{(2n+1)!}=\dfrac52 \sum_{n=0}^{\infty}\dfrac1{(2n)!}-\dfrac32\sum_{n=0}^{\infty}\dfrac1{(2n+1)!}$$

Now we know $\displaystyle\sum_{r=0}^\infty\dfrac{y^r}{r!}=e^y$

Can you find $e^y+e^{-y}$ and $e^y-e^{-y}$

Here $y=1$

Answer 2

Let $$(5n+1)=A(2n+1)+B \rightarrow 2A=5, A+B=1 \rightarrow $A=5/2, B=-3/2.$$ Then $$ S=\frac{5}{2} \sum_{n=0}^{\infty} \frac{1}{2n!} -\frac{3}{2} \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \frac{5}{2} \cosh 1 - \frac{3}{2} \sinh 1. $$