Prove $\sum_{i=1}^{n}\frac{i}{(i+1)!}= 1-\frac1{(n+1)!}$

Question

Required to prove: $1-\frac1{(k+2)!}$

$$\begin{align*} \sum_{i=1}^{k+1}\frac{i}{(i+1)!}&=\sum_{i=1}^k\frac{i}{(i+1)!}+\frac{k+1}{(k+2)!}\\ &= 1-\frac1{(k+1)!}+\frac{k+1}{(k+2)!}\tag{induction hyp.}\\[1em] \end{align*}$$

At my inductive solution i got stuck at the above section where i add the $k+1$ value to the inductive solution. Please help.

Answer 1

Let $$\sum_{i=1}^{k}\frac{i}{(i+1)!}=1-\frac{1}{(k+1)!}=\frac{(k+1)!-1}{(k+1)!}$$ we have $$\sum_{i=1}^{k+1}\frac{i}{(i+1)!}=\frac{k+1}{(k+2)!}+\sum_{i=1}^{k}\frac{i}{(i+1)!}$$ $$\qquad\qquad\qquad=\frac{k+1}{(k+2)!}+\frac{(k+1)!-1}{(k+1)!}$$ $$\qquad\qquad\qquad\quad=\frac{k+1+(k+2)!-(k+2)}{(k+2)!}$$ $$\quad=\frac{(k+2)!-1}{(k+2)!}$$ $$\quad=1-\frac{1}{(k+2)!}$$