Prove that $ \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+\cdots + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}$ for $n\in \mathbb N$

Question

I want to prove that if $n \in \mathbb N$ then $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}.$$

I think I am stuck on two fronts. First, I don't know how to express the leading terms on the left hand side before the $\dfrac{n}{(n+1)!}$ (or if doing so is even necessary to solve the problem). I am also assuming that the right high side should initially be expressed $1 - \dfrac{1}{(n+2)!}$. But where to go from there.

I'm actually not sure if I'm even thinking about it the right way.

Answer 1

Hint. If, for somespecific $n$, we have $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}\ ,$$ then $$\eqalign{\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n+1}{(n+2)!} &=1-\frac{1}{(n+1)!}+\frac{n+1}{(n+2)!}\cr &=1-\frac{n+2}{(n+2)!}+\frac{n+1}{(n+2)!}\cr &=1-\frac{1}{(n+2)!}\ .\cr}$$

Answer 2

We'll use induction to prove this

given statement is true for $n=1$

then we assume that it's true for $n=k$ $$\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{k}{(k+1)!} = 1 - \frac{1}{(k+1)!}$$ and then by adding $\dfrac{k+1}{(k+2)!}$ on both sides we get $$\begin{align} \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!}+ \cdots+ \frac{k}{(k+1)!} +\frac{k+1}{(k+2)!}&= 1 - \frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}\\ &=1 - \frac{k+2}{(k+2)!}+\frac{k+1}{(k+2)!}\\ &=1 +\frac{k+1-k-2}{(k+2)!}\\ &=1-\frac{1}{(k+2)!}\end{align}$$ it's also true for $n=k+1$ And then using Proncipal of mathematical induction...

Answer 3

Induction.

1. Base. $n = 1: \frac{1}{2!} = 1 - \frac{1}{2!}$.

2. Step. $n = m$ $-$ true. Let's prove for $m + 1$: $$ \frac{1}{2!} + \dots + \frac{m}{(m+1)!} + \frac{m+1}{(m+2)!} = 1 - \frac{1}{(m+1)!} = 1 - \frac{m+2}{(m+2)!} + \frac{m+1}{(m+2)!} = $$ $$ = 1 - \frac{1}{(m+2)!} $$

Answer 4

Hint: $\dfrac{n}{(n+1)!} = \dfrac{1}{n!} - \dfrac{1}{(n+1)!}$