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 \[\displaystyle\sum_{n=0}^{\infty}a_n=\sqrt{\dfrac{e\pi}{2}}\mathrm{erf}\left(\dfrac{1}{\sqrt{2}}\right)+\sqrt{e}\]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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 :

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

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

Nice ! +1

Log in to reply

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

Log in to reply