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 \(a_n=\dfrac{a_{n-2}}{n}\). Also, \(a_0=a_1=1\). Prove that \[\displaystyle\sum_{n=0}^{\infty}a_n=\sqrt{\dfrac{e\pi}{2}}\mathrm{erf}\left(\dfrac{1}{\sqrt{2}}\right)+\sqrt{e}\]

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## Comments

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TopNewestMy method is the method of generating functions :

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

Remember our recurrence relation is :

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

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

\( \displaystyle f(x) = { a }_{ 0 }+{ a }_{ 1 }x+\sum _{ n=2 }^{ \infty }{ { a }_{ n }{ x }^{ n } } \)

Since \( {a}_{0} = {a}_{1} =1 \)

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

Differentiating both sides with respect to \(x\) we have :

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

Using our recurrence relation we have :

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

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

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

\(\displaystyle y{ e }^{ -{ x }^{ 2 }/2 }=\int _{ 0 }^{ x }{ { e }^{ -{ t }^{ 2 }/2 }dt } + C \)

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

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

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

\( \displaystyle \sum _{ n=0 }^{ \infty }{ { a }_{ n } } =\sqrt { e } \int _{ 0 }^{ 1 }{ { e }^{ -{ t }^{ 2 }/2 }dt } +\sqrt { e } \)

With a little change of variables we have :

\( \displaystyle \sum _{ n=0 }^{ \infty }{ { a }_{ n } } =\sqrt { 2e } \int _{ 0 }^{ 1/\sqrt { 2 } }{ { e }^{ -{ x }^{ 2 } }dx } +\sqrt { e } \)

Using the defintion of error function we have finally :

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

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

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

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

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

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Nice ! +1

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

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