Proof of a Series for Pi
By Brad Rodgers (P1930) on Sunday, April 22, 2001 - 10:22 pm :

How would one prove



ó
õ 		1

0 
 x2n
1+x2
 dx= p
4
 - n
å
k=1 
 (-1)k+1
2k-1


Thanks,

Brad
By Kerwin Hui (Kwkh2) on Monday, April 23, 2001 - 03:17 pm :
Brad,

I think you missed out a minus sign somewhere. If you let In be ò01 x2n/(1+x2)dx, then

I0 = p/4

For n > 0, we have


In
=
ó
õ 		1

0 
 x2n/(1+x2)dx
=
ó
õ 		1

0 
 x2(n-1)(x2/(1+x2))dx
=
ó
õ 		1

0 
 x2(n-1)(1-1/(1+x2))dx
=
ó
õ 		1

0 
 x2(n-1)dx+ ó
õ 		1

0 
 x2(n-1)/(1+x2)dx
=
1/(2n-1)-In-1
Hence (inductively)

LHS=In = 1/(2n-1)-1/(2n-3)+¼+(-1)n +(-1)n+1p/4.

which is not what you have written on the RHS.

Kerwin

By Brad Rodgers (P1930) on Tuesday, April 24, 2001 - 03:56 am :

I think I forgot to put an absolute value in the RHS.

Thanks,

Brad