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The series is $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} ... + \frac{n-1}{n!},$$ which I can write as the sum $$\sum_{i=2}^n \frac{i-1}{i!}$$ I will try to evaluate the partial sums to look for a pattern. The first partial sums I get are $\frac12, \frac56 , \frac{23}{24}$. Thus, I make the guess that the sum is equal to $\frac{n!-1}{n!}$. I have already proven the base cases through experimentation, so I work on proving the induction step. I assume that $\sum_{i=2}^n \frac{i-1}{i!}= \frac{n!-1}{n!}$, for some $n=k$ so I get $$\sum_{i=2}^k \frac{i-1}{i!}= \frac{k!-1}{k!}$$ Now I prove this holds for $n=k+1$ as well. I get $$\sum_{i=2}^{k+1} \frac{i-1}{i!}= \frac{k!-1}{k!} + \frac{k}{(k+1)!} = \frac{(k!-1)(k+1)}{(k+1)!} + \frac{k}{(k+1)!} = \frac{(k+1)!-k-1}{(k+1)!} + \frac{k}{(k+1)!} = \frac{(k+1)!-1}{(k+1)!} $$ Thus the induction step holds as well, proving my expression is indeed the sum. However, this problem was in the telescoping sums section of my mathbook, so I was wondering how you would solve this sum telescopically since these induction proofs take some time.

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We have:

$$\sum_{i=2}^{n}\frac{i - 1}{i!} = \sum_{i=2}^{n}\frac{i}{i!} -\frac{1}{i!} = \sum_{i=2}^{n}\frac{1}{(i-1)!} - \frac{1}{i!}$$

$$=\bigg(\frac{1}{1!} - \frac{1}{2!}\bigg)+\bigg(\frac{1}{2!} - \frac{1}{3!}\bigg) +\bigg(\frac{1}{3!}-\frac{1}{4!}\bigg)+\dots+\bigg(\frac{1}{(n-1)!}-\frac{1}{n!}\bigg)$$

Telescoping:

$$=\frac{1}{1!}-\frac{1}{n!} = 1- \frac{1}{n!} = \boxed{\frac{n!-1}{n!}}$$

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Hint

$$\frac{n-1}{n!}= \frac{1}{(n-1)!}- \frac{1}{n!}$$ and therefore

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

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