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?

Question

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: $$1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1-\frac{1}{(k+2)!}$$

Answer 1

I thought it might be instructive to present a way forward that facilitates the analysis. To that end, note that we can write

$$\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$$

Thus, we have a telescoping sum and can write

$$\sum_{k=1}^n \frac{k}{(k+1)!}=\sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)=1-\frac{1}{(n+1)!}$$

where using induction on the sum as written $\sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)$ makes the induction proof almost trivial.

Answer 2

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.