Factorial Proof by Induction Question: $ \frac1{2!} + \frac2{3!} + \dots+ \frac{n}{(n+1)!} = 1 - \frac1{(n+1)!} $?

Question

$\text{Use the PMI to prove the following for all natural numbers n.}$

$ \frac{1}{2!} + \frac{2}{3!} + \cdot \cdot \cdot + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!} $

So for this question I get stuck because I cannot seem to make the left side and right side be equivalent.

$\color{Green}{Proof} :$

$\mathbf (I) \; Basis \; Step :$ Show $p(1).$ For $n=1$

$\frac{1}{(1+1)!} = 1- \frac{1}{(1+1)!} \Rightarrow \frac12 = \frac12$

$\mathbf (II) \; Inductive \; Step : $

Assume $p(k)$ for a PAC $k \in ℕ ,\; \; \;\; n=k$

$ \frac{1}{2!} + \frac{2}{3!} +\cdot\cdot\cdot + \frac{k}{(k+1)!} = 1-\frac{1}{(k+1)!} \; \; \text{Induction Hypothesis}$

Left Hand Side : We need to show : $p(k+1)$ i.e. $n = k+1$

$\frac{k}{(k+1)!} + \frac{k+1}{(k+1+1)!} \Rightarrow \ $

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

Right Hand Side: $ n =k+1$

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

This is where I have made my mistake because I think I have made an arithmetic mistake on the left side because I cannot get them to be verifiable. Any hints on how to correct the error would be appreciated. Have a good one.

Answer 1

By the inductive hypothesis, we know that \begin{align*} \frac{1}{2!} + \cdots + \frac{k}{(k+1)!} + \frac{k+1}{(k+2)!} &= 1 - \frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} \\ &= 1 - \frac{k+2}{(k+2)!} + \frac{k+1}{(k+2)!} \\ &= 1 - \frac{1}{(k+2)!}.\end{align*} Hence, if the statement is true for $n=k$, then it true for $n = k+1$, so by induction, it is true for all $n \in \mathbb{N}$.

Answer 2

What you need to show in the induction step is that

$$\frac1{2!}+\frac2{3!}+\ldots+\frac{k}{(k+1)!}+\frac{k+1}{(k+2)!}=1-\frac1{(k+2)!}\;.$$

Since your induction hypothesis is that the sum of the first $k$ terms on the left is $1-\frac1{(k+1)!}$, this amounts to showing that

$$1-\frac1{(k+1)!}+\frac{k+1}{(k+2)!}=1-\frac1{(k+2)!}\;.$$

If you multiply $\frac1{(k+1)!}$ by $1$ in the carefully chosen form $\frac{k+2}{k+2}$, you should have little difficulty showing this.

Answer 3

Although this post does not address the specific issue posed in the OP, I thought it might be instructive to present an alternative way forward. To that end, we note that

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

Now, the sum becomes a telescoping sum and we have immediately

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

as was to be shown!