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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.

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3 Answers 3

up vote 13 down vote accepted


$$\dfrac{k}{(k+1)!} = \dfrac{k+1-1}{(k+1)!} = \dfrac1{k!} - \dfrac1{(k+1)!}$$ 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, $$\lim_{n \to \infty} S_n = 1 - \lim_{n \to \infty} \dfrac1{(n+1)!} = 1$$

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Thanks you very much – user31280 Nov 7 '12 at 22:22
Complete solution to a homework type problem in only 1 minute! – GEdgar Nov 7 '12 at 22:40
One can also divide it into two sums of $e-1$ and $e-2$. I will add it to the answers but this stays has the chosen answer. – user31280 Nov 7 '12 at 22:52

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

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

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I like this approach – user31280 Nov 7 '12 at 22:30
@F'Ola: It’s not as elementary as Marvis’s, which I like, but it’s often quite useful. – Brian M. Scott Nov 7 '12 at 22:32
Actually, it is because for the past few days, I have been working with power series a lot and I should have remembered to use this one quicker than Marvis'. Thank you. – user31280 Nov 7 '12 at 22:36
@F'Ola: You’re welcome. – Brian M. Scott Nov 7 '12 at 22:38

Just a remix of Marvis' solution

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

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