(from zhangwuji) $$ex sumlimits_{n=0}^{infty}dfrac{n^3+2n+1}{(n^4+n^2+1)n!},quad sumlimits_{n=0}^{infty}dfrac{1}{(n^4+n^2+1)n!}. eex$$
解答: (by wangsb) $$eex ea sumlimits_{n=0}^{infty}dfrac{n^3+2n+1}{(n^4+n^2+1)n!} =&sumlimits_{n=0}^{infty}dfrac{n^3+2n+1}{(n^2+n+1)(n^2-n+1)n!}\ =&sumlimits_{n=0}^{infty}Big(frac{n}{n^2+n+1}+frac{1}{n^2-n+1}Big)frac{1}{n!}\ =&sumlimits_{n=0}^{infty}dfrac{1}{(n^2-n+1)n!}+sumlimits_{n=1}^{infty}dfrac{n}{((n+1)^2-(n+1)+1)n!}\ =&sumlimits_{n=0}^{infty}dfrac{1}{(n^2-n+1)n!}+sumlimits_{n=2}^{infty}dfrac{n-1}{(n^2-n+1)(n-1)!}\ =&sumlimits_{n=0}^{infty}dfrac{1}{(n^2-n+1)n!}+sumlimits_{n=2}^{infty}dfrac{n(n-1)}{(n^2-n+1)n!}\ =&1+1+sumlimits_{n=2}^{infty}dfrac{1+n(n-1)}{(n^2-n+1)n!}=sumlimits_{n=0}^{infty}dfrac{1}{n!}=e. eea eeex$$ $$eex ea sumlimits_{n=0}^{infty}dfrac{1}{(n^4+n^2+1)n!}=&sumlimits_{n=0}^{infty}dfrac{1}{n!}-sumlimits_{n=0}^{infty}dfrac{n^4+n^2}{(n^4+n^2+1)n!}\ =&e-frac{1}{2}sumlimits_{n=1}^{infty}frac{n}{(n-1)!}Big(dfrac{1}{n^2+n+1}+frac{1}{n^2-n+1}Big)\ =&e-frac{1}{2}sumlimits_{n=1}^{infty}frac{n^2}{n!(n+2+n+1)}\ &-frac{1}{2}sumlimits_{n=1}^{infty}frac{n}{(n-1)!((n-1)^2+(n-1)+1)}\ =&e-frac{1}{2}sumlimits_{n=1}^{infty}frac{n^2}{n!(n+2+n+1)}-frac{1}{2}sumlimits_{n=0}^{infty}frac{n+1}{n!(n^2+n+1)}\ =&e-frac{1}{2}-frac{1}{2}sumlimits_{n=1}^{infty}frac{n^2+n+1}{n!(n^2+n+1)}=e-frac{e}{2}=frac{e}{2}. eea eeex$$