<s>[INST] I am stuck on proving $\frac1{2!}+\frac2{3!}+\dots+\frac{n}{(n+1)!}=1-\frac1{(n+1)!}$ by induction, could anyone check my work? I will skip the Base Case step.
This is the questions.

Use mathematical induction to prove that$$\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$for all integers $n\ge 1$.

This is my proof:
$$\sum_{i=1}^n \frac{i}{(i+1)!} = 1-\frac{1}{(n+1)!}$$
$$\sum_{i=1}^k \frac{i}{(i+1)!} = 1-\frac{1}{(k+1)!}$$
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!} = 1-\frac{1}{(k+2)!}$$
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!} = \sum_{i=1}^k \frac{i}{(i+1)!} + \frac{k+1}{(k+2)!}$$
$$\sum_{i=1}^{k+1} \frac{i}{(i+1)!} = 1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!}$$
And this is where I am stuck, I don't know how to prove that: $$1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1-\frac{1}{(k+2)!}$$
 [/INST] Note that $(k + 2)! =(k + 2) \cdot (k + 1)! $. You then get for your last equation $$ 1 - \frac{1}{(k + 1)!} + \frac{k + 1}{(k + 2)!} = 1 - \frac{k + 2}{(k + 2)!} + \frac{k + 1}{(k + 2)!} = 1 + \frac{1 - 2 + k - k}{(k + 2)!} = 1 - \frac{1}{(k + 2)!}$$
which proves the statement.
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