Evaluate lim_(n->oo) 1/n^4 prod_(j=1)^(2n) (n^2+j^2)^(1/n)?

2 Answers
Jun 29, 2017

lim_(n->oo) 1/n^4 prod_(j=1)^(2n) (n^2+j^2)^(1/n) = 25e^( 2arctan(2)-4)

Explanation:

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)

Jun 29, 2017

5^2e^(2 arctan2-4)

Explanation:

Calling m = 2n

lim_(n->oo) 1/n^4 prod_(j=1)^(2n) (n^2+j^2)^(1/n) = lim_(m->oo)(2/m)^4 prod_(k=1)^m (1+(k/(m/2))^2)^(2/m)((m/2)^2)^(2/m) =

= lim_(m->oo)prod_(k=1)^m(1+((2k)/m)^2)^(2/m) = y

Now

log y=lim_(m->oo)2/m sum_(k=1)^m log(1+((2k)/m)^2)

Calling xi = k/m we have

log y = 2 int_0^1 log(1+(2xi)^2) d xi

but

int log(1+(2xi)^2) d xi = 2 (arctan(2 xi) + xi log(1 + 4 xi^2)-2 xi)+C

then

log y = 2 (arctan2 + Log5-2) and finally

y = e^(2 (arctan2 + Log5-2))=5^2e^(2 arctan2-4)