Prove via induction $\sum_{k=2}^{n}{\frac{k-1}{k!}} = \frac{n!-1}{n!}, \forall n \in \mathbb{N}, n \ge 2$

Question

I have to prove by induction, that

$$\sum_{k=2}^{n}{\frac{k-1}{k!}} = \frac{n!-1}{n!}, \forall n \in \mathbb{N}, n \ge 2$$

$\begin{align} \sum_{k=2}^{n+1}{\frac{k-1}{k!}} &= \sum_{k=2}^{n}{\frac{k-1}{k!} + \frac{(n+1)-1}{(n+1)!}} \\ &= \sum_{k=2}^{n}{\frac{k-1}{k!} + \frac{n}{(n+1)!}} \\ &= \frac{n!-1}{n!} + \frac{n}{(n+1)!} \\ &= \frac{[n!-1](n+1)!}{n!(n+1)!} + \frac{n! \cdot n}{n!(n+1)!} \\ &= \frac{[n!-1](n+1)! + n! \cdot n}{n!(n+1)!} \\ &= \frac{n!(n+1)!-(n+1)!+n!n}{n!(n+1)!} \\ \end{align}$

Question: How should I go on ? Do I have a mistake until now?

Just telescope: $\frac{k-1}{k!}= \frac{1}{(k-1)!}-\frac{1}{k!}$ — mathworker21, Jan 24 '18 at 17:03
Hint: The common denominator of $\frac 1{n!}$ and $\frac 1{(n+1)!}$ is much smaller — Hagen von Eitzen, Jan 24 '18 at 17:03
You are overcomplicating it: $$\frac{k-1}{k!} = \frac{1}{(k-1)!}-\frac{1}{k!}$$ is a telescopic term, hence the final outcome ($1-\frac{1}{n!}$) is clear. — Jack D'Aurizio, Jan 24 '18 at 17:04
@JaideepKhare: same story, $$ \left(1-\frac{1}{n!}\right)+\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right) = 1-\frac{1}{(n+1)!}.$$ — Jack D'Aurizio, Jan 24 '18 at 17:07

score 3 · Accepted Answer · answered Jan 24 '18 at 17:04

From your third line:

\begin{align} \frac{n!-1}{n!} + \frac{n}{(n+1)!} &= \frac{(n!-1)(n+1)}{n!(n+1)}+\frac{n}{(n+1)!} \\ &= \frac{(n+1)!-(n+1)+n}{(n+1)!} \\ &=\frac{(n+1)!-1}{(n+1)!} \end{align}

Don't forget to check base case.

To go on from where you stop, divide numerator and denominator by $n!$.

score 3 · Answer 2 · answered Jan 24 '18 at 17:08

3

Perfect up to this point; to continue, write the middle $(n+1)!$ in the numerator as $(n+1)n!$, then cancel the $n!$ from the denominator.

answered Jan 24 '18 at 17:08

BallBoy

14,472

Prove via induction $\sum_{k=2}^{n}{\frac{k-1}{k!}} = \frac{n!-1}{n!}, \forall n \in \mathbb{N}, n \ge 2$

2 Answers2