[再寄小读者之数学篇](2014-06-03 计算两个无穷级数)

(from zhangwuji) $$\bex \sum\limits_{n=0}^{\infty}\dfrac{n^3+2n+1}{(n^4+n^2+1)n!},\quad \sum\limits_{n=0}^{\infty}\dfrac{1}{(n^4+n^2+1)n!}. \eex$$

解答: (by wangsb) $$\beex \bea \sum\limits_{n=0}^{\infty}\dfrac{n^3+2n+1}{(n^4+n^2+1)n!} =&\sum\limits_{n=0}^{\infty}\dfrac{n^3+2n+1}{(n^2+n+1)(n^2-n+1)n!}\\ =&\sum\limits_{n=0}^{\infty}\Big(\frac{n}{n^2+n+1}+\frac{1}{n^2-n+1}\Big)\frac{1}{n!}\\ =&\sum\limits_{n=0}^{\infty}\dfrac{1}{(n^2-n+1)n!}+\sum\limits_{n=1}^{\infty}\dfrac{n}{((n+1)^2-(n+1)+1)n!}\\ =&\sum\limits_{n=0}^{\infty}\dfrac{1}{(n^2-n+1)n!}+\sum\limits_{n=2}^{\infty}\dfrac{n-1}{(n^2-n+1)(n-1)!}\\ =&\sum\limits_{n=0}^{\infty}\dfrac{1}{(n^2-n+1)n!}+\sum\limits_{n=2}^{\infty}\dfrac{n(n-1)}{(n^2-n+1)n!}\\ =&1+1+\sum\limits_{n=2}^{\infty}\dfrac{1+n(n-1)}{(n^2-n+1)n!}=\sum\limits_{n=0}^{\infty}\dfrac{1}{n!}=e. \eea \eeex$$ $$\beex \bea \sum\limits_{n=0}^{\infty}\dfrac{1}{(n^4+n^2+1)n!}=&\sum\limits_{n=0}^{\infty}\dfrac{1}{n!}-\sum\limits_{n=0}^{\infty}\dfrac{n^4+n^2}{(n^4+n^2+1)n!}\\ =&e-\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{n}{(n-1)!}\Big(\dfrac{1}{n^2+n+1}+\frac{1}{n^2-n+1}\Big)\\ =&e-\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{n^2}{n!(n+2+n+1)}\\ &-\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{n}{(n-1)!((n-1)^2+(n-1)+1)}\\ =&e-\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{n^2}{n!(n+2+n+1)}-\frac{1}{2}\sum\limits_{n=0}^{\infty}\frac{n+1}{n!(n^2+n+1)}\\ =&e-\frac{1}{2}-\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{n^2+n+1}{n!(n^2+n+1)}=e-\frac{e}{2}=\frac{e}{2}. \eea \eeex$$