<s>[INST] Prove by induction: $\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}$ 
Prove
  $$\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}.$$

My problem with this is that it doesn't hold for the base case: $n=1$. This question is from the book "Abstract Algebra" by Charles Pinter (Page 212, exercise 7). When I go about proving it, I can't reach my goal which is $\frac{(n+1)!-1}{(n+1)!}=1-\frac{1}{(n+1)!}$, for $n=n+1$.
 [/INST] $$\frac{1}{2!}+\frac{2}{3!}+...+\frac{n}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!} $$ 
Name as $p(n)$
1 step:$$n=1 \to p(1):\frac{1}{2!}=\frac{2!-1}{2!} $$
2nd step :assume $$n=k \to p(k):\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}=\frac{(k+1)!-1}{(k+1)!} $$ 
3rd step :prove for n=k+1 $$n=k+1 \to p(k+1)=\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}+\frac{k+1}{(k+2)!}=\frac{(k+2)!-1}{(k+2)!}$$ so substitute $\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}={\color{Red}{\frac{(k+1)!-1}{(k+1)!} }}$ into p(k+1) 
$${\color{Red}{\frac{(k+1)!-1}{(k+1)!} }}+\frac{k+1}{(k+2)!}=\\\frac{(k+1)!-1}{(k+1)!}*\frac{k+2}{k+2}+\frac{k+1}{(k+2)!} =\\\frac{(k+1)!(k+2)-1(k+2)}{(k+2)!}+\frac{k+1}{(k+2)!}=\\ \frac{(k+1)!(k+2)-1(k+2)+(k+1)}{(k+2)!} =\\\frac{(k+1)!(k+2)-1}{(k+2)!} =\\\frac{(k+2)!-1}{(k+2)!} $$ the proof is complete now (by induction)
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