Sum

Calculus Level 3

∑ k = 1 ∞ k − 1 k ! = ? \large \sum_{k=1}^\infty \frac{k-1}{k!} = \, ?

3 5.7 1 2.5

Abdullah Mughal by Abdullah Mughal , 18, Switzerland

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1 solution

∑ k = 1 ∞ k − 1 k ! = ∑ k = 2 ∞ k − 1 k ! Since k − 1 = 0 = ∑ k = 1 ∞ k + 1 − 1 ( k + 1 ) ! = ∑ k = 1 ∞ ( 1 k ! − 1 ( k + 1 ) ! ) = ∑ k = 1 ∞ 1 k ! − ∑ k = 2 ∞ 1 k ! = 1 1 ! = 1 \begin{aligned} \sum_{\color{#3D99F6}k=1}^\infty \frac {k-1}{k!} & = \sum_{\color{#D61F06}k=2}^\infty \frac {k-1}{k!} & \small \color{#D61F06} \text{Since }k-1 = 0 \\ & = \sum_{\color{#3D99F6}k=1} ^\infty \frac {{\color{#3D99F6}k+1}-1}{({\color{#3D99F6}k+1})!} \\ & = \sum_{k=1} ^\infty \left(\frac 1{k!} -\frac 1{(k+1)!} \right) \\ & = \sum_{k=1} ^\infty \frac 1{k!} - \sum_{k=2} ^\infty \frac 1{k!} \\ & = \frac 1{1!} = \boxed{1} \end{aligned}

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Nice solution sir .

Sabhrant Sachan - 4 years, 5 months ago

Sir, kindly explain the 2nd and 3rd statement (especially the 2nd).

A Former Brilliant Member - 3 years, 5 months ago

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k + 1 − 1 ( k + 1 ) ! = k + 1 ( k + 1 ) ! − 1 ( k + 1 ) ! = k + 1 ( k + 1 ) k ! − 1 ( k + 1 ) ! = 1 k ! − 1 ( k + 1 ) ! \dfrac {k+1-1}{(k+1)!} = \dfrac {k+1}{(k+1)!} - \dfrac 1{(k+1)!} = \dfrac {k+1}{(k+1)k!} - \dfrac 1{(k+1)!} = \dfrac 1{k!} - \dfrac 1{(k+1)!}

Chew-Seong Cheong - 3 years, 5 months ago

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No, Sir. I understood that much. Notice that you used "k=2" under the summation symbol. How did you do it?

A Former Brilliant Member - 3 years, 5 months ago

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@A Former Brilliant Member – Expand ∑ k = 1 ∞ k − 1 k ! = 0 1 ! + 1 2 ! + 2 3 ! + 3 4 ! + ⋯ = 0 + ∑ k = 2 ∞ k − 1 k ! \displaystyle \sum_{\color{#3D99F6}k=1}^\infty \frac {k-1}{k!} = \frac 0{1!} + \frac 1{2!} + \frac 2{3!} + \frac 3{4!} + \cdots = 0 + \sum_{\color{#D61F06}k=2}^\infty \frac {k-1}{k!}

Chew-Seong Cheong - 3 years, 5 months ago

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@Chew-Seong Cheong – Thank you for your explanation

A Former Brilliant Member - 3 years, 5 months ago

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