证明下列级数收敛:

 

(1). $\dps{\vsm{n}\sez{\frac{1}{n}-\ln\sex{1+\frac{1}{n}}}}$;

 

(2). $\dps{\vsm{n}\sez{e-\sex{1+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}}}}$. (东北师范大学)

 

证明: 由 Taylor 展开, $$\bex \ln (1+x)=x-\frac{1}{2(1+\xi_x)^2} x^2,\quad e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\frac{e^\eta}{(n+1)!}x^{n+1}. \eex$$ 而 $$\beex \bea 0<\frac{1}{n}-\ln\sex{1+\frac{1}{n}}=\frac{1}{2(1+\xi_n)^2}\frac{1}{n^2}<\frac{1}{2n^2},\\ 0<e-\sex{1+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}}=\frac{e^\eta}{(n+1)!} <\frac{e}{(n+1)n}. \eea \eeex$$ 故所讨论级数均收敛.