<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] Try this in your inductive step:
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!}= \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+2)!} $$
Notice the denominator is $(k + 2)!$ instead of $(k+1)!$, which makes simplification easier:
$$ = \frac{1}{(k+1)(k-1)!} + \frac{k+1}{(k+2)!}  $$
$$ = \frac{(k + 2)k}{(k+2)!} + \frac{k+1}{(k+2)!} = \frac{k^2 + 3k + 1}{(k+2)!} $$ 
$$ = \frac{(k + 2)(k + 1) - 1}{(k+2)!} = \frac{1}{k!} - \frac{1}{(k+2)!}$$
It looks like your inductive hypothesis is incorrect.
</s>