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 
$$ \sum_{i=1}^{k+1} \frac{i}{(i+1)!} = \frac{1}{(k+2)(k)!} $$
 A: 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.
A: Let's see what
that sum actually is,
since your result is
obviously wrong.
$\begin{array}\\
\sum_{i=1}^n \frac{i}{(i+1)!}
&=\sum_{i=1}^n \frac{i+1-1}{(i+1)!}\\
&=\sum_{i=1}^n \frac{i+1}{(i+1)!}-\sum_{i=1}^n \frac{1}{(i+1)!}\\
&=\sum_{i=1}^n \frac{1}{i!}-\sum_{i=2}^{n+1} \frac{1}{i!}\\
&=(1+\sum_{i=2}^n \frac{1}{i!})-(\sum_{i=2}^{n} \frac{1}{i!}+\frac1{(n+1)!})\\
&=1-\frac1{(n+1)!}\\
\end{array}
$
And that is what
you should try to prove
by induction.
A: 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.
