Find the sum of the series

Calculus Level 5

1 + 1 2 + 2 2 2 ! + 1 2 + 2 2 + 3 2 3 ! + 1 2 + 2 2 + 3 2 + 4 2 4 ! + \large 1+\frac{1^2+2^2}{2!} + \frac{1^2+2^2+3^2}{3!} + \frac{1^2+2^2+3^2+4^2}{4!} + \cdots

If the value of the sum above can be expressed in the form of a e b \dfrac{ae}{b} , where a a and b b are coprime positive integers, find a + b a+b .


Clarification: e 2.71828 e \approx 2.71828 denotes the Euler's number .


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The answer is 23.

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2 solutions

Guilherme Niedu
May 15, 2017

We know that:

e x = n = 0 x n n ! \large \displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}

Differentiating with respect to x x and multiplying by x x three times one gets:

x e x = n = 0 n x n n ! \large \displaystyle xe^x = \sum_{n=0}^{\infty} \frac{nx^n}{n!}

( x 2 + x ) e x = n = 0 n 2 x n n ! \large \displaystyle (x^2+x)e^x = \sum_{n=0}^{\infty} \frac{n^2x^n}{n!}

( x 3 + 3 x 2 + x ) e x = n = 0 n 3 x n n ! \large \displaystyle (x^3+3x^2+x)e^x = \sum_{n=0}^{\infty} \frac{n^3x^n}{n!}

Making x = 1 x=1 :

e = n = 0 n n ! \color{#20A900} \boxed{ \large \displaystyle e = \sum_{n=0}^{\infty} \frac{n}{n!} } 2 e = n = 0 n 2 n ! \color{#20A900} \boxed{ \large \displaystyle 2e = \sum_{n=0}^{\infty} \frac{n^2}{n!} } 5 e = n = 0 n 3 n ! \color{#20A900} \boxed{ \large \displaystyle 5e = \sum_{n=0}^{\infty} \frac{n^3}{n!} }

Then:

S = n = 1 k = 1 n k 2 n ! \large \displaystyle S = \sum_{n=1}^{\infty} \sum_{k=1}^n \frac{k^2}{n!}

S = 1 6 n = 1 n ( n + 1 ) ( 2 n + 1 ) n ! \large \displaystyle S = \frac16 \sum_{n=1}^{\infty} \frac{n(n+1)(2n+1)}{n!}

S = 1 6 n = 0 n ( n + 1 ) ( 2 n + 1 ) n ! \large \displaystyle S = \frac16 \sum_{n=0}^{\infty} \frac{n(n+1)(2n+1)}{n!}

S = 1 6 [ 2 n = 0 n 3 n ! + 3 n = 0 n 2 n ! + n = 0 n n ! ] \large \displaystyle S = \frac16 \left [ 2 \sum_{n=0}^{\infty} \frac{n^3}{n!} + 3\sum_{n=0}^{\infty} \frac{n^2}{n!} + \sum_{n=0}^{\infty} \frac{n}{n!} \right ]

S = 1 6 [ 2 5 e + 3 2 e + e ] \large \displaystyle S = \frac16 \left [ 2 \cdot 5e + 3 \cdot 2e + e \right]

S = 17 e 6 \color{#20A900} \boxed{ \large \displaystyle S = \frac{17e}{6} }

Thus:

a = 17 , b = 6 , a + b = 23 \color{#3D99F6} \large \displaystyle a=17, b = 6, \boxed{a+b=23}

WOWWWW.BEAUTIFUL SOLUTION.

rajdeep brahma - 4 years ago

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Thanks, sir!

Guilherme Niedu - 4 years ago

This is bashing. I have forgotten a little but it might be neatly done with exponential generating function. Maybe someone can help here.

Kartik Sharma - 4 years ago

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Yes... Using expansion series for e x e^x and differentiating 3 3 times with appropriate manipulations gives n = 1 n ( n + 1 ) ( 2 n + 1 ) x n 1 2 6 n ! = e x 3 ( x 5 2 + 9 x 3 2 2 + 3 x 1 2 ) \sum_{n=1}^{\infty}\dfrac{n(n+1)(2n+1)x^{n-\frac 12}}{6n!}=\dfrac{e^x}3\left(x^{\frac 52}+\frac{9x^\frac 32}{2}+3x^{\frac 12}\right) Just substituting x = 1 x=1 gives the result.

Rishabh Jain - 4 years ago

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Yeah maybe I was mistaken. That is not really "neat". I was thinking if there was a method to find

n = 1 k = 1 n k 2 1 n ! \displaystyle \sum_{n=1}^\infty{\sum_{k=1}^n{k^2}\frac{1}{n!}}

as a probable product of two functions. Just like we are able to find n = 1 k = 1 n k 2 x n = 1 1 x n = 1 n 2 x n \displaystyle \sum_{n=1}^\infty {\sum_{k=1}^n{k^2} x^n} = \frac{1}{1-x} \sum_{n=1}^\infty{n^2 x^n}

I will check out.

Kartik Sharma - 4 years ago

Super cool dude. I have done this by making a series of summision in e. I just love your solution.

Nivedit Jain - 3 years, 12 months ago
Nivedit Jain
Jun 13, 2017

Nice solution!!

Viraam Rao - 3 years, 5 months ago

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