Question :
Find limit : $$\lim_{n\rightarrow＋\infty}\sum_{k＝1}^{n}\frac {1}{k(k＋1)(k＋1)!}$$
My attemp:
$$\lim_{n\rightarrow＋\infty}\sum_{k＝1}^{n}\frac {1}{k(k＋1)(k＋1)!}=\lim_{n\rightarrow＋\infty}\sum_{k＝1}^{n}\left(\frac {1}{k}－\frac {1}{k＋1}\right)\cdot\frac {1}{(k＋1)!}$$
I appreciate your help
Best Answer
Following Robert's Hint $$\frac{1}{k(k＋1)(k＋1)!}=\frac{(k+1)^2-k-k(k+1)}{k(k＋1)(k＋1)!}=\frac {1}{kk!}-\frac {1}{(k＋1)(k+1)!}-\frac {1}{(k+1)!}$$ So our question becomes $$\sum_{k=1}^{\infty}\frac {1}{kk!}-\frac {1}{(k＋1)(k+1)!}-\frac {1}{(k+1)!}$$ You can see that the first two terms telescope so we're left with $$1-\sum_{k=1}^{\infty}\frac{1}{(k+1)!}$$ $$=1-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\right)$$ Recall that $$e=\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\right)$$ So $$1-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\right)=1-(e-2)$$ $$\boxed{=3-e}$$