Inspired by Ikkyu San

Calculus Level 5

∑ n = − 1 ∞ 1 ( n 2 ) ! \large \sum _{ n=-1 }^{ \infty }{ \frac { 1 }{ \left( \frac { n }{ 2 } \right) ! } }

The summation above can be expressed as 1 π + e a ( erf ( b ) + c ) , \dfrac1{\sqrt { \pi }} +{ e }^{ a }\left( \text{erf}\left( b \right) +c \right) , where a , b , c a,b,c are integers.

Find a + b + c a+b+c .


The answer is 3.

Joel Yip by Joel Yip , 21, Singapore

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1 solution

Aman Rajput
May 20, 2021

There is slight error in the question, we wanted to split the sum as ∑ n = − 1 ∞ 1 ( n / 2 ) ! = 1 ( − 1 / 2 ) ! + ∑ n = 0 ∞ 1 ( n / 2 ) ! \sum_{n=-1}^{\infty}\frac{1}{(n/2)!}=\frac{1}{(-1/2)!}+\sum_{n=0}^{\infty}\frac{1}{(n/2)!} = 1 π + e + e E r f ( 1 ) =\frac{1}{\sqrt{\pi}} + e+eErf(1)

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