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.
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.
HINT
$\dfrac{k}{(k+1)!} = \dfrac{k+1-1}{(k+1)!} = \dfrac1{k!} - \dfrac1{(k+1)!}$ and make use of telescopic cancellation.
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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, $\lim_{n \to \infty} S_n = 1 - \lim_{n \to \infty} \dfrac1{(n+1)!} = 1$
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
$\sum_{k\ge 1}\frac{k}{(k+1)!}=1\;.$
Just a remix of Marvis' solution
$\begin{align*}\sum_{k =1}^{n}\dfrac{k}{(k+1)!} &=\sum_{k =1}^{n}\left( \dfrac1{k!} - \dfrac1{(k+1)!}\right)\\&= \sum_{k =1}^{n} \dfrac1{k!}-\sum_{k =1}^{n}\dfrac1{(k+1)!}\\&= \sum_{k =1}^{n} \dfrac1{k!}-\sum_{k =2}^{n}\dfrac1{k!}\\ &=\sum_{k =0}^{n} \dfrac1{k!}-1-\left(\sum_{k =0}^{n}\dfrac1{k!}-1-1\right) \\ &=e-1-(e-2)\\&=1\\ \end{align*}$