How would you prove that $\sum_{k=1}^{n}\frac{k}{(k+1)!}= \frac{(n+1)!-1}{(n+1)!}$

Question

And another task from an old exam:

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

for all $n \in \mathbb{N}$.

I smell proof by induction but I don't know how to do the start. I know how an induction proof is made up and done but I cannot really apply it on this task. Firstly, the sum symbol seems annoying for me, very annoying.

But I did the start by putting $k=1$ on the left side and $n=1$ on the right side:

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

$$\sum_{k=1}^{1}\frac{1}{2} = \frac{1}{2}$$

which is true.

So if it works for one $n$, it should work for $n+1$ as well (here is another problem, shall I concentrate on the left or the right side..?!).

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

Hmm and from this point, also when it starts getting exciting, I didn't feel well continuing because... yeah the sum symbol is very confusing here.

Okay, maybe a little change that might look it a bit shorter / more nice:

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

Did I do it correctly till here at all? Is my approach of using induction correct...?

Answer 1

Since $k=(k+1)-1$ and $(k+1)!=(k+1)\cdot k!$ we have $$\sum_{k=1}^{n}\frac{k}{(k+1)!} = \sum_{k=1}^{n}\frac{1}{k!}-\sum_{k=1}^{n}\frac{1}{(k+1)!} = \color{red}{\frac{1}{1!}-\frac{1}{(n+1)!}}$$ i.e. a telescopic sum.

Answer 2

It is always time for an overkill. Since $$\sum_{k=0}^{n+1}\frac{x^{k}}{k!}=\frac{e^{x}\Gamma\left(n+2,x\right)}{\left(n+1\right)!} $$ where $\Gamma\left(a,b\right) $ is the incomplete Gamma function, follows that $$\sum_{k=2}^{n+1}\frac{x^{k-1}}{k!}=\frac{e^{x}\Gamma\left(n+2,x\right)}{\left(n+1\right)!x}-\frac{1}{x}-1 $$ so taking the derivative $$\sum_{k=2}^{n+1}\frac{\left(k-1\right)x^{k-2}}{k!}=\frac{e^{x}\left(x-1\right)\Gamma\left(n+2,x\right)-x^{n+2}}{\left(n+1\right)!x^{2}}+\frac{1}{x^{2}} $$ and the result follows taking $x=1 $ $$\sum_{k=2}^{n+1}\frac{\left(k-1\right)}{k!}=\sum_{k=1}^{n}\frac{k}{\left(k+1\right)!}=\color{red}{1-\frac{1}{\left(n+1\right)!}}.$$