Who could have done it?

Calculus Level 4

1 λ = 2 2 9801 ∑ k = 0 ∞ ( 4 k ) ! ( 1103 + 26390 k ) ( k ! ) 4 39 6 4 k \dfrac{1}{\lambda} = \dfrac{2\sqrt{2}}{9801}\displaystyle\sum^\infty_{k=0} \dfrac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}

Solve for the value of λ \lambda .


The answer is 3.141.

Siddharth Bhatt by siddharth bhatt , 25, India

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2 solutions

Kunal Gupta
Jun 20, 2015

Ramanujan!! :P

1 Helpful 0 Interesting 0 Brilliant 1 Confused

bro Ramanujan is not the answer. Please provide a solution

Shashank Rustagi - 5 years, 11 months ago

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I would love to see you provide a solutiom shashank.

Isaac Buckley - 5 years, 11 months ago
Axelrod Polaris
Jul 12, 2015

Ramanujan found this as an infinite series for 1/pi. But even if you don't know that, testing values for k quickly shows that the denominator is expanding much faster than the top, so quickly that rounding to 3 places is already 0 when k = 1. So just plug in k = 0 to get (9801/2*sqrt(2)) * 1/1103 = pi.

0 Helpful 0 Interesting 0 Brilliant 1 Confused

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