The Numerators Diverge Too

Calculus Level 2

It is well known that

e = ∑ k = 0 ∞ 1 k ! = 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + ⋯ e = \sum_{k=0}^\infty \dfrac{1}{k!} = \dfrac{1}{ 0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots

What is the value of

∑ k = 0 ∞ k + 1 k ! = 1 0 ! + 2 1 ! + 3 2 ! + 4 3 ! + ⋯ ? \sum_{k=0}^{\infty} \dfrac{ k+1} {k!} = \dfrac{1}{0!} + \dfrac{2}{ 1!} + \dfrac{3}{2!} + \dfrac{ 4}{3!} + \cdots \quad ?

Notation : ! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

Bonus : Prove this by just manipulating the terms from the first identity.

( k + 1 ) e (k+1) e 2 e 2e e + 2 e + 2 e 2 e^2

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Calvin Lin by Calvin Lin

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5 solutions

Pi Han Goh
Aug 24, 2016

By the ratio test , we can show that the series in question converges. This is because a n = n + 1 n ! ⇔ a n + 1 = n + 2 ( n + 1 ) ! ⇒ a n + 1 a n = n + 2 ( n + 1 ) 2 → 0 . a_n = \dfrac{n+1}{n!} \; \Leftrightarrow \; a_{n+1} = \dfrac{n+2}{(n+1)!} \; \Rightarrow \; \dfrac{a_{n+1}}{a_n} = \dfrac{n+2}{(n+1)^2} \to 0 \; .

So we can perform arithmetic on the desired series

∑ k = 0 ∞ k + 1 k ! − ∑ k = 0 ∞ 1 k ! = ∑ k = 0 ∞ k + 1 − 1 k ! = ∑ k = 0 ∞ k k ! = 0 0 ! + ∑ k = 1 ∞ k k ! = ∑ k = 1 ∞ 1 ( k − 1 ) ! = ∑ k = 0 ∞ 1 k ! ( ∑ k = 0 ∞ k + 1 k ! ) − e = e ∑ k = 0 ∞ k + 1 k ! = 2 e . \begin{aligned} \sum_{k=0}^\infty \dfrac{k+1}{k!} - \sum_{k=0}^\infty \dfrac{1}{k!} &=& \sum_{k=0}^\infty \dfrac{k+1-1}{k!} = \sum_{k=0}^\infty \dfrac{k}{k!} = \dfrac{0}{0!} + \sum_{k=1}^\infty \dfrac k{k!} = \sum_{k=1}^\infty \dfrac 1{(k-1)!} = \sum_{k=0}^\infty \dfrac 1{k!} \\ \left(\sum_{k=0}^\infty \dfrac{k+1}{k!}\right) - e &=& e \\ \sum_{k=0}^\infty \dfrac{k+1}{k!} &=& \boxed{2e} \; . \end{aligned}

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Michael Mendrin
Aug 22, 2016

Okay, the summation can be first split

∑ k = 0 ∞ k + 1 k ! = ∑ k = 0 ∞ 1 ( k − 1 ) ! + ∑ k = 0 ∞ 1 k ! = 1 ( 0 − 1 ) ! + 2 ∑ k = 0 ∞ 1 k ! \displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { k+1 }{ k! } } =\displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { 1 }{ (k-1)! } } +\displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { 1 }{ k! } } =\dfrac { 1 }{ (0-1)! } +2\displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { 1 }{ k! } }

Since 1 ( 0 − 1 ) ! = 0 \dfrac { 1 }{ (0-1)! }=0 , the answer immediately follows But this is the non-trivial hiccup part. In a way, by summing the series expansion for e e with itself shifted over by one term, we can prove that this hiccup is true.

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Deleted my previous solution but thought of one now. So, I am using the comment here.

e x = ∑ k = 0 ∞ x k k ! x e x = ∑ k = 0 ∞ x k + 1 k ! d d x x e x = d d x ∑ k = 0 ∞ x k + 1 k ! e x + x e x = ∑ k = 0 ∞ ( k + 1 ) x k k ! Putting x = 1 2 e = ∑ k = 0 ∞ k + 1 k ! \begin{aligned} e^x & = \sum_{k=0}^\infty \frac {x^k}{k!} \\ xe^x & = \sum_{k=0}^\infty \frac {x^{k+1}}{k!} \\ \frac d{dx} xe^x & = \frac d{dx} \sum_{k=0}^\infty \frac {x^{k+1}}{k!} \\ e^x + xe^x & = \sum_{k=0}^\infty \frac {(k+1)x^k}{k!} & \small \color{#3D99F6}{\text{Putting }x=1} \\ 2e & = \sum_{k=0}^\infty \frac {k+1}{k!} \end{aligned}

Chew-Seong Cheong - 4 years, 9 months ago

Hm, only issue is that it's not ideal to work with 1 ( − 1 ) ! \frac{ 1}{ (-1)!} . Instead, we have 0 0 ! \frac{0}{0!} , which is clearly 0.

Calvin Lin Staff - 4 years, 9 months ago

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Calvin, the quicker way to prove the value of your summation, using your suggestion, " just manipulating the terms from the first identity", is to note that

1 n ! + 1 ( n + 1 ) ! = n + 2 ( n + 1 ) ! \dfrac { 1 }{ n! } +\dfrac { 1 }{ (n+1)! } =\dfrac { n+2 }{ (n+1)! }

but, hey, I was having fun----I'm trying to say that this would be a neat way to prove that the following has to be true necessarily

1 ( 0 − 1 ) ! = 0 \dfrac { 1 }{ (0-1)! }=0

Addendum: Here goes the proper proof of your summation

∑ k = 0 ∞ 1 k ! = e \displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { 1 }{ k! } } =e

1 + ∑ k = 0 ∞ 1 ( k + 1 ) ! = e 1+\displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { 1 }{ (k+1)! } } =e

Hence

1 + ∑ k = 0 ∞ 1 k ! + ∑ k = 0 ∞ 1 ( k + 1 ) ! = 1 + ∑ k = 0 ∞ k + 2 ( k + 1 ) ! = ∑ k = 0 ∞ k + 1 k ! = 2 e 1+\displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { 1 }{ k! } + \displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { 1 }{ (k+1)! } } } =1+\displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { k+2 }{ (k+1)! } = } \displaystyle \sum _{ k=0 }^{ \infty }{ \dfrac { k+1 }{ k! } } =2e

.

Michael Mendrin - 4 years, 9 months ago

Could we simply ignore the negative factorial terms ? Don't they have some meaning ?

Anurag Pandey - 4 years, 9 months ago

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That's always a bad idea, "ignoring things because they don't make sense".

Michael Mendrin - 4 years, 8 months ago

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@Michael Mendrin – The how should I think around it ?

Anurag Pandey - 4 years, 8 months ago

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@Anurag Pandey – Like my first answer to Calvin Lin. 1 ( − 1 ) ! \dfrac{1}{\left(-1\right)!} really does work out to 0 0 , and THEN you can ignore it!

Here's the plot of the function 1 x ! \dfrac{1}{x!}

Michael Mendrin - 4 years, 8 months ago

∑ k = 0 ∞ k + 1 k ! = ∑ k = 0 ∞ k k ! + ∑ k = 0 ∞ 1 k ! \displaystyle \sum_{k=0}^{\infty} \frac{k+1}{k!} =\displaystyle \sum_{k=0}^{\infty}\frac{k}{k!} +\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!} = 0 1 ! + ∑ k = 1 ∞ 1 ( k − 1 ) ! + e = ∑ k = 0 ∞ 1 k ! + e = 2 e =\frac{0}{1!}+\displaystyle \sum_{k=1}^{\infty} \frac{1}{(k-1)!} +e=\displaystyle \sum_{k=0}^{\infty} \frac{1}{k!} +e=2e

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Mark Recio
Sep 26, 2016

Since

e = 1 0 ! + 1 1 ! + 1 2 ! + . . . e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + ...

= 1 1 ! + 2 2 ! + 3 3 ! + . . . \frac{1}{1!} + \frac{2}{2!} + \frac{3}{3!} + ...

= 1 0 ! + 2 1 ! + 3 2 ! + . . . − ( 1 0 ! + 1 1 ! + 1 2 ! + . . . ) \frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + ... - ( \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + ... )

=> 1 0 ! + 2 1 ! + 3 2 ! + . . . = e + e = 2 e \frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + ... = e + e = 2e .

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Shaun Lee
Sep 7, 2016

k + 1 k ! \frac{k+1}{k!} =(k+1)÷k!= k k ! \frac{k}{k!} +e= 1 ( k − 1 ) ! \frac{1}{(k-1)!} +e=2e

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