How to compute $\sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$? I don't know how to compute $\sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$,appreciate any help!
Is there any general rule for solving such problems?

$$\text{As }  e^x=\sum_{0\le r<\infty}\frac{x^r}{r!} $$
 $$\text{and }\frac{(2n+1)}{n!}x^{2n+1}=2x^3 \cdot \frac{(x^2)^{n-1}}{(n-1)!}+x\cdot \frac{(x^2)^n}{n!}$$
$$\text{So,} \sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$$
$$=2x^3 \cdot \sum_{n=0}^{\infty}\frac{(x^2)^{n-1}}{(n-1)!}+x\cdot \sum_{n=0}^{\infty}\frac{(x^2)^n}{n!}$$
$$=2x^3 \cdot \sum_{m=0}^{\infty}\frac{(x^2)^m}{m!}+x\cdot \sum_{n=0}^{\infty}\frac{(x^2)^n}{n!}\text{ putting }  n-1=m  \text{ and using} \frac1{(-1)!}=0 $$
$$=2x^3 \cdot e^{x^2}+x\cdot e^{x^2}$$
