Prove that $\sum_{k\ge 1} k/(k+1)! = 1$

Question

Prove that $$\sum_{k\ge 1} \cfrac k{(k+1)!} = 1$$

I know it is one because it is the sum to infinity of a probability distribution but I can't prove it.

Answer 1

HINT

$$\dfrac{k}{(k+1)!} = \dfrac{k+1-1}{(k+1)!} = \dfrac1{k!} - \dfrac1{(k+1)!}$$ and make use of telescopic cancellation.

Move your mouse over the gray area for the complete solution.

Let $$S_n = \sum_{k =1}^{n} \dfrac{k}{(k+1)!}$$ From the hint, we then have \begin{align} S_n & = \left(\dfrac1{1!} - \dfrac1{2!} \right) + \left(\dfrac1{2!} - \dfrac1{3!} \right) + \left(\dfrac1{3!} - \dfrac1{4!} \right) + \cdots + \left(\dfrac1{(n-1)!} - \dfrac1{n!} \right) + \left(\dfrac1{n!} - \dfrac1{(n+1)!} \right)\\ & = 1 - \dfrac1{(n+1)!} \end{align} Hence, $$\lim_{n \to \infty} S_n = 1 - \lim_{n \to \infty} \dfrac1{(n+1)!} = 1$$

Answer 2

Start with the Maclaurin series $$e^x=\sum_{k\ge 0}\frac{x^k}{k!}\;.$$ Subtract $1$ from both sides and factor out an $x$ on the right to get

$$e^x-1=x\sum_{k\ge 1}\frac{x^{k-1}}{k!}\;.$$

Now differentiate both sides with respect to $x$:

$$\begin{align*} e^x&=\sum_{k\ge 1}\frac{x^{k-1}}{k!}+x\sum_{k\ge 1}\frac{(k-1)x^{k-2}}{k!}\\ &=\frac{e^x-1}x+x\sum_{k\ge 0}\frac{kx^{k-1}}{(k+1)!}\\ &=\frac{e^x-1}x+x\sum_{k\ge 1}\frac{kx^{k-1}}{(k+1)!}\;. \end{align*}$$

Finally, rearrange to get

$$\sum_{k\ge 1}\frac{kx^{k-1}}{(k+1)!}=\frac1x\left(e^x-\frac{e^x-1}x\right)$$

and evaluate at $x=1$ to get

$$\sum_{k\ge 1}\frac{k}{(k+1)!}=1\;.$$

Answer 3

Just a remix of Marvis' solution

$$\begin{align*}\sum_{k =1}^{n}\dfrac{k}{(k+1)!} &=\sum_{k =1}^{n}\left( \dfrac1{k!} - \dfrac1{(k+1)!}\right)\\&= \sum_{k =1}^{n} \dfrac1{k!}-\sum_{k =1}^{n}\dfrac1{(k+1)!}\\&= \sum_{k =1}^{n} \dfrac1{k!}-\sum_{k =2}^{n}\dfrac1{k!}\\ &=\sum_{k =0}^{n} \dfrac1{k!}-1-\left(\sum_{k =0}^{n}\dfrac1{k!}-1-1\right) \\ &=e-1-(e-2)\\&=1\\ \end{align*}$$