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

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Well, how would you add fractions with different denominators? – Hagen von Eitzen Mar 11 at 6:33
Try multiplying the second term on the left hand side of your last equation by $\frac{k+2}{k+2}$ and use the definition of factorial. – Troy Mar 11 at 6:33
I got $$1-\frac{k+2}{(k+2)!} + \frac{k+1}{(k+2)!}$$ – 00111000 Mar 11 at 6:36
BTW there are probably some other posts on this site about the same sum. For example:… – Martin Sleziak Mar 11 at 8:32
up vote 6 down vote accepted

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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Ya, your method worked – 00111000 Mar 11 at 6:45

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


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.

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you write "that does not rely on induction" - how do you prove the statement on telescopic sums without induction? – Steven Mar 11 at 6:46
@Steven Relying on a theorem the proof of which might or might not rely on induction, is not the same as a direct proof by induction. – Dr. MV Jun 11 at 16:30

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