1. How to solve Ramanujan's problem by numerical method and We need to find I0
I0 =
1
1
2
1
3
1
4
1 ...



Define that; I1 =
1
1
3
1
4
1
5
1 ...



And; I2 =
1
1
4
1
5
1
6
1 ...



In the general case; In =
1
1 n 2( ) I
n 1
 Or; In+1 =
1 I
n

n 2( ) I
n

Notice I0 that 0 < I0 < 1
So, We need to find the condition of Ik+1 in the Programming1 when k is the large number
Initial Condition for Programming k 9999
Programming 1; Assume I0 = 0.5
FindValue1 k( ) I
0
0.5
I
n 1
1 I
n

n 2( ) I
n


n 0 kfor
I

I FindValue1 k( )
0
0
1
2
3
4
5
0.5
0.5
0.333
0.5
0.2
...

So that; I
k 1
0.01
We can approximate that; I
k 1
0
Programming 2;
FindValue2 k( ) I
k 1
0
I
n
1
1 n 2( ) I
n 1


n k 0for
I

I FindValue2 k( )
0
0
1
2
3
4
5
0.525
0.452
0.404
0.369
0.342
...

So that, the finally; I
0
0.52513527616098121