A calculus problem by Hamza A

Calculus Level 5

∑ n = 0 ∞ 1 ( n 2 ) ! = e ( a + erf ( b ) ) \large \displaystyle\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ \left( \frac { n }{ 2 } \right) ! } } =e(a+\text{ erf }(b))

The equation above holds true for positive integers a a and b b . Find a + b a+b .


The answer is 2.

Hamza A by Hamza A , 19, USA

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1 solution

Joel Yip
Apr 14, 2016

Firstly nice problem.

Rearranging, ∑ n = 0 ∞ 1 ( n 2 ) ! = 1 ( 0 2 ) ! + 1 ( 1 2 ) ! + 1 ( 2 2 ) ! + ⋯ = 1 0 ! + 1 ( 1 2 ) ! + 1 1 ! + 1 ( 1 + 1 2 ) ! + ⋯ = 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + ⋯ + 1 ( 1 2 ) ! + 1 ( 1 + 1 2 ) ! + 1 ( 2 + 1 2 ) ! + ⋯ = ∑ n = 0 ∞ 1 n ! + ∑ m = 0 ∞ 1 ( m + 1 2 ) ! \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ \left( \frac { n }{ 2 } \right) ! } } =\frac { 1 }{ \left( \frac { 0 }{ 2 } \right) ! } +\frac { 1 }{ \left( \frac { 1 }{ 2 } \right) ! } +\frac { 1 }{ \left( \frac { 2 }{ 2 } \right) ! } +\cdots \\ =\frac { 1 }{ 0! } +\frac { 1 }{ \left( \frac { 1 }{ 2 } \right) ! } +\frac { 1 }{ 1! } +\frac { 1 }{ \left( 1+\frac { 1 }{ 2 } \right) ! } +\cdots \\ =\frac { 1 }{ 0! } +\frac { 1 }{ 1! } +\frac { 1 }{ 2! } +\frac { 1 }{ 3! } +\dots +\frac { 1 }{ \left( \frac { 1 }{ 2 } \right) ! } +\frac { 1 }{ \left( 1+\frac { 1 }{ 2 } \right) ! } +\frac { 1 }{ \left( 2+\frac { 1 }{ 2 } \right) ! } +\cdots \\ =\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } } +\sum _{ m=0 }^{ \infty }{ \frac { 1 }{ \left( m+\frac { 1 }{ 2 } \right) ! } }

The left sum is well-known and the sum would be e e . The proof is here .

The right sum is ∑ m = 0 ∞ 1 ( m + 1 2 ) ! = 1 ( 1 2 ) ! + 1 ( 1 + 1 2 ) ! + 1 ( 2 + 1 2 ) ! + ⋯ \sum _{ m=0 }^{ \infty }{ \frac { 1 }{ \left( m+\frac { 1 }{ 2 } \right) ! } } =\frac { 1 }{ \left( \frac { 1 }{ 2 } \right) ! } +\frac { 1 }{ \left( 1+\frac { 1 }{ 2 } \right) ! } +\frac { 1 }{ \left( 2+\frac { 1 }{ 2 } \right) ! } +\cdots

Now if we use gamma functions, Γ ( n + 1 ) = n ! \Gamma \left( n+1 \right) =n!

1 ( 1 2 ) ! + 1 ( 1 + 1 2 ) ! + 1 ( 2 + 1 2 ) ! + ⋯ = 1 Γ ( 3 2 ) + 1 Γ ( 5 2 ) + 1 Γ ( 7 2 ) + ⋯ = 1 1 2 Γ ( 1 2 ) + 1 3 2 Γ ( 3 2 ) + 1 5 2 Γ ( 5 2 ) = 1 1 2 × Γ ( 1 2 ) + 1 1 2 × 3 2 × Γ ( 1 2 ) + 1 1 2 × 3 2 × 5 2 × Γ ( 1 2 ) + ⋯ \frac { 1 }{ \left( \frac { 1 }{ 2 } \right) ! } +\frac { 1 }{ \left( 1+\frac { 1 }{ 2 } \right) ! } +\frac { 1 }{ \left( 2+\frac { 1 }{ 2 } \right) ! } +\cdots =\frac { 1 }{ \Gamma \left( \frac { 3 }{ 2 } \right) } +\frac { 1 }{ \Gamma \left( \frac { 5 }{ 2 } \right) } +\frac { 1 }{ \Gamma \left( \frac { 7 }{ 2 } \right) } +\cdots \\ =\frac { 1 }{ \frac { 1 }{ 2 } \Gamma \left( \frac { 1 }{ 2 } \right) } +\frac { 1 }{ \frac { 3 }{ 2 } \Gamma \left( \frac { 3 }{ 2 } \right) } +\frac { 1 }{ \frac { 5 }{ 2 } \Gamma \left( \frac { 5 }{ 2 } \right) } \\ =\frac { 1 }{ \frac { 1 }{ 2 } \times \Gamma \left( \frac { 1 }{ 2 } \right) } +\frac { 1 }{ \frac { 1 }{ 2 } \times \frac { 3 }{ 2 } \times \Gamma \left( \frac { 1 }{ 2 } \right) } +\frac { 1 }{ \frac { 1 }{ 2 } \times \frac { 3 }{ 2 } \times \frac { 5 }{ 2 } \times \Gamma \left( \frac { 1 }{ 2 } \right) } +\cdots

Using the Euler's Reflection Formula,

Γ ( 1 2 ) Γ ( 1 − 1 2 ) = π s i n ( π 2 ) = π \Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( 1-\frac { 1 }{ 2 } \right) =\frac { \pi }{ sin\left( \frac { \pi }{ 2 } \right) } \\ =\pi

Γ ( 1 2 ) 2 = π Γ ( 1 2 ) = π { \Gamma \left( \frac { 1 }{ 2 } \right) }^{ 2 }=\pi \\ \Gamma \left( \frac { 1 }{ 2 } \right) =\sqrt { \pi }

So the sum becomes this,

1 π ( 1 1 2 + 1 1 2 × 3 2 + 1 1 2 × 3 2 × 5 2 + ⋯ ) \frac { 1 }{ \sqrt { \pi } } \left( \frac { 1 }{ \frac { 1 }{ 2 } } +\frac { 1 }{ \frac { 1 }{ 2 } \times \frac { 3 }{ 2 } } +\frac { 1 }{ \frac { 1 }{ 2 } \times \frac { 3 }{ 2 } \times \frac { 5 }{ 2 } } +\cdots \right)

1 π ( 1 1 2 + 1 1 2 × 3 2 + 1 1 2 × 3 2 × 5 2 + ⋯ ) = 1 π ( 2 1 1 + 2 2 1 × 3 + 2 3 1 × 3 × 5 + ⋯ ) = 1 π ∑ n = 0 ∞ 2 2 n + 1 ( 2 n + 1 ) ! ! \frac { 1 }{ \sqrt { \pi } } \left( \frac { 1 }{ \frac { 1 }{ 2 } } +\frac { 1 }{ \frac { 1 }{ 2 } \times \frac { 3 }{ 2 } } +\frac { 1 }{ \frac { 1 }{ 2 } \times \frac { 3 }{ 2 } \times \frac { 5 }{ 2 } } +\cdots \right) =\frac { 1 }{ \sqrt { \pi } } \left( \frac { { 2 }^{ 1 } }{ 1 } +\frac { { 2 }^{ 2 } }{ 1\times 3 } +\frac { { 2 }^{ 3 } }{ 1\times 3\times 5 } +\cdots \right) \\ =\frac { 1 }{ \sqrt { \pi } } \sum _{ n=0 }^{ \infty }{ \frac { { 2 }^{ 2n+1 } }{ \left( 2n+1 \right) !! } }

e r f ( x ) = e − x 2 π ∑ n = 0 ∞ ( 2 x ) 2 n + 1 ( 2 n + 1 ) ! ! erf\left( x \right) =\frac{e^{-x^2}}{\sqrt{\pi}} \sum_{n=0}^\infty {\frac{(2x)^{2n+1}}{(2n+1)!!}}

So this sum is e × e r f ( 1 ) e\times erf\left( 1 \right)

Not forgeting to add the earlier e e ,

makes e + e × e r f ( 1 ) = e ( 1 + e r f ( 1 ) ) e+e\times erf\left( 1 \right) =e\left( 1+erf\left( 1 \right) \right)

so 1 + 1 = 2 1+1=2 .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Great explanation of this problem. Do we know what e r f ( 1 ) erf (1) is equal to?

thanks!

nice solution btw :)

Hamza A - 5 years, 2 months ago

1 π ∫ − 1 1 e − x 2 d x \frac { 1 }{ \sqrt { \pi } } \int _{ -1 }^{ 1 }{ { e }^{ -{ x }^{ 2 } } } dx

Joel Yip - 5 years, 2 months ago

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