The document describes a numerical method to solve Ramanujan's problem to calculate the value of J0. It defines a series of terms I0, I1, I2, etc and J0, J1, J2, etc and provides recurrence relations to calculate successive terms. It then presents 4 programming solutions to calculate the values of these terms up to a large value of k using the recurrence relations. The results of the 4 programs show that as k increases, Ik approaches 0.525 and Jk approaches 1.431175463899617.

1. How to solve Ramanujan's problem by numerical method and We need to find J0
J0 =
1
1
2
1
3
1 ...


2
1
3
1
4
1 ...


1
1
2
1
3
1 ...


3
1
4
1
5
1 ...


1
1
2
1
3
1 ...


...
Define that; I0 =
1
1
2
1
3
1 ...


I1 =
1
1
3
1
4
1 ...


1
1
4
1
5
1 ...

And; I2 =
In the general case; In =
1
1 n 2( ) I
n 1
 Or; In+1 =
1 I
n

n 2( ) I
n

Then; J0 = I
0
2 I
1
 I
0
3 I
2
I
0
4 I
3
I
0
...
Define that; J1 = I
0
3 I
2
 I
0
4 I
3
I
0
4 I
3
I
0
...
And; J2 = I
0
4 I
3
 I
0
5 I
4
I
0
6 I
5
I
0
...
In the general case; Jn = I
0
n 2( ) I
n 1
 I
0
n 3( ) I
n 2
 I
0
n 4( ) I
n 3
 I
0
...
Jn = I
0
n 2( ) I
n 1
 J
n 1

Or; Jn+1 =
J
n 
2
I
0

n 2( ) I
n 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

2. 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
0.525
Programming 3; Assume J0 = 0.75
FindValue3 k( ) J
0
0.75
J
n 1
J
n 
2
I
0

n 2( ) I
n 1


n 0 k 1for
J

J FindValue3 k( )
0
0
1
2
3
4
5
0.75
0.041
-0.432
-0.229
-0.276
...

So that; J
k
2.493 10
5

We can approximate that; J
k
0

3. Programming 4;
FindValue4 k( ) J
k
0
J
n
I
0
n 2( ) I
n 1
 J
n 1

n k 1 0for
J

J FindValue4 k( )
0
0
1
2
3
4
5
1.431
1.684
1.908
2.111
2.298
...

So that, the finally; J
0
1.431175463899617