For every #n#, there are #2n# factors in the product. We can then split the common factor #1/n^4# among them:
#1/n^4 = prod_(j=1)^(2n) (1/n^4)^(1/(2n)) = prod_(j=1)^(2n) 1/(n^2)^(1/n)#
Then we have:
#a_n = 1/n^4 prod_(j=1)^(2n) (n^2+j^2)^(1/n) = prod_(j=1)^(2n) (n^2+j^2)^(1/n) / (n^2)^(1/n) = prod_(j=1)^(2n) (1+j^2/n^2)^(1/n)#
Pose now:
#b_n = ln(a_n) = ln(prod_(j=1)^(2n) (1+j^2/n^2)^(1/n))#
Based on the properties of logarithms:
#b_n = sum_(j=1)^(2n) 1/n ln (1+j^2/n^2)#
Write the sum as:
#b_n = 2 sum_(j=1)^(2n) 1/(2n) ln (1+(2j)^2/(2n)^2)#
Consider now the function:
#f(x) = ln(1+4x^2)#
over the interval #x in (0, 1)#.
As:
#f'(x) = (2x)/(1+x^2) > 0 # in the interval, the function is strictly increasing.
Partition the interval in #n# sub-intervals of length #1/(2n)#
#(0,1) = uu_(j=1)^(2n) ( ( j-1)/(2n), j/(2n))#
and build the Riemann sum over this partition. As the function is increasing, the minimum value is on the left boundary of the interval, and the maximum is on the right boundary:
#s_n = sum_(j=1)^(2n) (1/(2n))ln (1+ (4(j-1)^2)/(2n^2)) #
#S_n = sum_(j=1)^(2n) (1/(2n))ln (1+ (4j^2)/(2n)^2) =1/2 sum_(j=1)^(2n) b_n#
Thus, if the integral converges we have:
#lim_(n->oo) b_n = 2 int_0^1 ln(1+4x^2)dx#
Solve the integral by parts:
# int_0^1 ln(1+4x^2)dx = [xln(1+4x^2)dx]_0^1 - 8int_0^1 x^2/(1+4x^2)dx#
# int_0^1 ln(1+4x^2)dx = ln5- 2int_0^1 (1+4x^2-1)/(1+4x^2)dx#
# int_0^1 ln(1+4x^2)dx = ln5 - 2int_0^1 dx+ 2int_0^1 1/(1+4x^2)dx#
# int_0^1 ln(1+4x^2)dx = ln5 - 2[x]_0^1+ int_0^1 1/(1+(2x)^2)d(2x)#
# int_0^1 ln(1+4x^2)dx = ln5 -2 + [arctan(2x)]_0^1#
# int_0^1 ln(1+4x^2)dx = ln5 -2 + arctan(2)#
So we have:
#lim_(n->oo) b_n = 2ln5 -4 +2arctan(2)#
As #e^x# is uniformly continuous for #x in RR#:
#lim_(n->oo) a_n = lim_(n->oo) e^(b_n) = e^((lim_(n->oo) b_n))#
Then:
#lim_(n->oo) a_n = e^(2ln5 -4 +2arctan(2))#
ans simplifying:
#lim_(n->oo) a_n = 25e^( 2arctan(2)-4)#
