A Recursive Sum

I was playing around with recursions and I found something interesting which I would like to share with you guys. Do give it a try.

We recursively define an=an−2na_n=\dfrac{a_{n-2}}{n}. Also, a0=a1=1a_0=a_1=1. Prove that ∑n=0∞an=eπ2erf(12)+e\displaystyle\sum_{n=0}^{\infty}a_n=\sqrt{\dfrac{e\pi}{2}}\mathrm{erf}\left(\dfrac{1}{\sqrt{2}}\right)+\sqrt{e}

#Calculus #RecurrenceRelations #Summation #DifferentialEquations

Note by Pratik Shastri
6 years, 3 months ago

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Comments

My method is the method of generating functions :

y=f(x)=∑n=0∞anxn\displaystyle y=f(x)=\sum _{ n=0 }^{ \infty }{ { a }_{ n }{ x }^{ n } }

Remember our recurrence relation is :

(n+2)an+2=an \displaystyle (n+2){ a }_{ n+2 }={ a }_{ n }

Also f(x)f(x) can be written as :

f(x)=a0+a1x+∑n=2∞anxn \displaystyle f(x) = { a }_{ 0 }+{ a }_{ 1 }x+\sum _{ n=2 }^{ \infty }{ { a }_{ n }{ x }^{ n } }

Since a0=a1=1 {a}_{0} = {a}_{1} =1

⇒f(x)=1+x+∑n=0∞an+2xn+2 \displaystyle \Rightarrow f(x) = 1+x+\sum _{ n=0 }^{ \infty }{ { a }_{ n+2 }{ x }^{ n+2 } }

Differentiating both sides with respect to xx we have :

f′(x)=1+∑n=0∞(n+2)an+2xn+1 \displaystyle f'(x)=1+\sum _{ n=0 }^{ \infty }{ (n+2){ a }_{ n+2 }{ x }^{ n+1 } }

Using our recurrence relation we have :

f′(x)=1+∑n=0∞anxn+1 \displaystyle f'(x)=1+\sum _{ n=0 }^{ \infty }{ { a }_{ n }{ x }^{ n+1 } }

⇒f′(x)=1+xf(x) \displaystyle \Rightarrow f'(x)=1+xf(x)

Now it's a linear differential equation and it's solution is as follows :

ye−x2/2=∫0xe−t2/2dt+C\displaystyle y{ e }^{ -{ x }^{ 2 }/2 }=\int _{ 0 }^{ x }{ { e }^{ -{ t }^{ 2 }/2 }dt } + C

Now at x=0,y=1x=0 , y=1 hence it comes out to be C=1 C=1

Finally y=f(x)=ex2/2∫0xe−t2/2dt+ex2/2 \displaystyle y = f(x) ={ e }^{ { x }^{ 2 }/2 }\int _{ 0 }^{ x }{ { e }^{ -{ t }^{ 2 }/2 }dt } +{ e }^{ { x }^{ 2 }/2 }

To get the value of summation simply put x=1x=1 to get :

∑n=0∞an=e∫01e−t2/2dt+e \displaystyle \sum _{ n=0 }^{ \infty }{ { a }_{ n } } =\sqrt { e } \int _{ 0 }^{ 1 }{ { e }^{ -{ t }^{ 2 }/2 }dt } +\sqrt { e }

With a little change of variables we have :

∑n=0∞an=2e∫01/2e−x2dx+e \displaystyle \sum _{ n=0 }^{ \infty }{ { a }_{ n } } =\sqrt { 2e } \int _{ 0 }^{ 1/\sqrt { 2 } }{ { e }^{ -{ x }^{ 2 } }dx } +\sqrt { e }

Using the defintion of error function we have finally :

∫0ze−x2dx=π2erf(z)\displaystyle \int _{ 0 }^{z}{ { e }^{ -{ x }^{ 2 } }dx } = \frac { \sqrt { \pi } }{ 2 } \mathrm{erf}(z)

∑n=0∞an=eπ2erf(12)+e \displaystyle \sum _{ n=0 }^{ \infty }{ { a }_{ n } } =\sqrt { \frac { e\pi }{ 2 } } \mathrm{erf}(\frac { 1 }{ \sqrt { 2 } } )+\sqrt { e }

Ronak Agarwal - 6 years, 3 months ago

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Cool! That's exactly what I did too! Ditto! +1!

Pratik Shastri - 6 years, 3 months ago

Nice ! +1

A Former Brilliant Member - 6 years, 3 months ago

Wonder if there is another way to get to the sum..

Pratik Shastri - 6 years, 3 months ago

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There is an another way you can solve for the general term and then sum can be calculated very easily.

Ronak Agarwal - 6 years, 3 months ago

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@Ronak Agarwal – The general term looks horrid. How do we sum it?

Pratik Shastri - 6 years, 3 months ago

Excellent solution as usual. And now thanks to You I learned how to solve infinite sums using generating functions.

Oussama Boussif - 6 years, 3 months ago
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