×

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

Note by Pratik Shastri
2 years, 3 months ago

Sort by:

My 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 :

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

$$\displaystyle \sum _{ n=0 }^{ \infty }{ { a }_{ n } } =\sqrt { \frac { e\pi }{ 2 } } \mathrm{erf}(\frac { 1 }{ \sqrt { 2 } } )+\sqrt { e }$$ · 2 years, 3 months ago

Excellent solution as usual. And now thanks to You I learned how to solve infinite sums using generating functions. · 2 years, 3 months ago

Wonder if there is another way to get to the sum.. · 2 years, 3 months ago

There is an another way you can solve for the general term and then sum can be calculated very easily. · 2 years, 3 months ago

The general term looks horrid. How do we sum it? · 2 years, 3 months ago

Nice ! +1 · 2 years, 3 months ago