∫du/√(1-u4), u∈[0,1]. This definite integral must tend to infinity because 1/√(1-u4) is not defined when u=1. The integral is the area beneath the curve between the given limits. u=1 is an asymptote so this area is infinite.
Series method:
1/√(1-u4)=(1-u4)-½=1+u4/2+(1)(3)u8/22+(1)(3)(5)u12/23+...+[(2n-1)(2n-3)...(3)(1)u4n/2n]+...
We can write the series as ∑(2n-1)(2n-3)...(3)(1)u4n/2n for n≥0. (n!=1 for n≤0)
This can be written ∑[(2n-1)!/(2n-1(n-1)!)]u4n/2n=∑(2n-1)!u4n/[22n-1(n-1)!]
Example: the 4th term (n=3) is (1)(3)(5)u12/23=15u12/8; using the formula: 5!u12/[25(2!)]=120u12/64=15u12/8.
So ∫∑(2n-1)!u4n/[22n-1(n-1)!]du=∑(2n-1)!u4n+1/[22n-1(n-1)!(4n+1)].
Example: n=3: 120u13/(64×13)=15u13/104 is the 4th term after integration.
Applying the limits we get:
∑(2n-1)!/[22n-1(n-1)!(4n+1)]=
1+1/10+1/12+15/104+105/272+45/32+2079/320+... which tends to infinity, because the terms get larger as the series progresses.