Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$

Question

Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}$

So I proved the base case where $n=1$ and got $\frac{1}{2}$

Then since $n=k$ implies $n=k+1$ I setup the problem like so:

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

After trying to simplify it I got the following:

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

However, I'm having trouble simplifying it to match the RHS. Hints?

Answer 1

The equality you want to prove is $$ \underbrace{\frac{1}{2!}+\dots+\frac{k}{(k+1)!}}_{*}+ \frac{k+1}{(k+2)!}=1-\frac{1}{(k+2)!} $$ The term marked $*$ is equal, by the induction hypothesis, to $$ 1-\frac{1}{(k+1)!} $$ and so you need to manipulate $$ 1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!} $$ Hint: $$ 1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!} = 1-\frac{k+2}{(k+2)!}+\frac{k+1}{(k+2)!} =\dots $$ Do the necessary steps in order to finish up at $1-\dfrac{k+1}{(k+2)!}$.

Answer 2

You have your base case.

Explicitly state your inductive hypothesis.

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

We will show that:

$\sum_\limits{n=1}^{k+1}\frac{n}{(n+1)!} = 1-\frac{1}{(k+2)!}$

$\sum_\limits{n=1}^k\frac{n}{(n+1)!}+\frac{(k+1)}{(k+2)!}$

$1-\frac{1}{(k+1)!}+\frac{(k+1)}{(k+2)!}$(By the inductive hypothesis.)

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

Answer 3

I thought it might be instructive to present an approach that is not a proof by induction. To that end, note that we can write

$$\bbox[5px,border:2px solid #C0A000]{\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}}$$

Therefore, we have a telescoping sum such that

$$\begin{align} \sum_{k=1}^n \frac{k}{(k+1)!}&=\sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)\\\\ &=1-\frac{1}{(n+1)!} \end{align}$$

And we are done!

Answer 4

You can simply write it like this also this solution doesn't need induction:

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

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

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

$$solved!$$