<s>[INST] Prove by induction, that $ \sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!}$ If I'm not wrong,
$$\sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!},$$
but I am having trouble proving it by induction...
If $n=1$ the formulas coincide.
Sup $n=k$ is valid: $$\sum_{i=1}^k \frac{i}{(i+1)!}= \frac{1}{(k+1)(k-1)!}.$$
Then if $n=k+1$,
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!}= \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+1)!} $$
$$ = \frac{1}{(k+1)(k-1)!} + \frac{k+1}{(k+1)!}  $$
$$ = \frac{k + (k +1)}{(k+1)!} = \frac{2k + 1}{(k+1)!} $$ I really don't see from this how I should arrive to 
$$ \sum_{i=1}^{k+1} \frac{i}{(i+1)!} = \frac{1}{(k+2)(k)!} $$
 [/INST] For $n=2$ we have
$$
\frac{1}{2!}+\frac{2}{3!}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}
$$
On the other hand,
$$
\frac{1}{(2+1)(2-1)!}=\frac{1}{3}
$$
so your conjecture is wrong.
For $n=3$ we have
$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}=
\frac{12}{4!}+\frac{8}{4!}+\frac{3}{4!}=
\frac{23}{24}
$$
and you can make a better conjecture.
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