I would like to prove the following proposition by induction:
For all $m\in\mathbb N$, we have $\sum_{i=1}^m \frac{i}{(i+1)!} = 1 - \frac{1}{(m+1)!}.$
First, I prove the base step $m = 1$:
$\sum_{i=1}^1 \frac{i}{(i+1)!}$ = 1 - $\frac{1}{(2)!}$, which is effectively $\frac{1}{2}$ on both sides of this equation.
Next, I assume that there is some $k\in\mathbb N$ such that $$ \sum_{i=1}^k \frac{i}{(i+1)!} = 1 - \frac{1}{(k+1)!} $$
Now, to prove the inductive step I do: $$ \sum_{i=1}^{k+1} = 1 - \frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} $$
At this point, I don't know what to do because I think I messed up at the last step. Any ideas?
You're almost there. From $1 - \frac{1}{(k+1)!} + \frac{k+1}{(k+2)!}$ note that $\frac{1}{(k+1)!} = \frac{k+2}{(k+2)!}$ and you can finish the proof.