It seems rational but is truly irrational

Calculus Level 4

n = 0 ( 4 n ) ! ( 1103 + 26390 n ) ( n ! ) 4 39 6 4 n \displaystyle \sum_{n = 0}^{\infty} \dfrac {(4n)! (1103 + 26390n)}{(n!)^4 396^{4n}}

has a value such that the closest integer to it is x x . If the actual value of it is a b c π \dfrac {a}{b \sqrt{c} \cdot \pi} where the fraction is in the lowest terms and c c is prime, find the value of x + a + b + c x + a + b + c ?


The answer is 10908.

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1 solution

Julian Poon
Nov 14, 2014

One definition of π \pi is: 1 π = 2 2 9801 n = 0 ( 4 n ) ! ( 1103 + 26390 n ) ( n ! ) 4 396 4 n \frac { 1 }{ \pi } =\frac { 2\sqrt { 2 } }{ 9801 } \sum _{ n=0 }^{ \infty }{ \frac { (4n)!(1103+26390n) }{ { (n!) }^{ 4 }{ 396 }^{ 4n } } } So with some simple algebra, we can get: n = 0 ( 4 n ) ! ( 1103 + 26390 n ) ( n ! ) 4 396 4 n = 9801 π × 2 2 \sum _{ n=0 }^{ \infty }{ \frac { (4n)!(1103+26390n) }{ { (n!) }^{ 4 }{ 396 }^{ 4n } } } =\frac { 9801 }{ \pi \times 2\sqrt { 2 } } which makes the answer 1103 + 9801 + 2 + 2 = 10908 1103+9801+2+2=\boxed{10908}

Do you have a solution that does not require knowing that?

Sharky Kesa - 6 years, 7 months ago

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Nope. sorry.

Julian Poon - 6 years, 7 months ago

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