The number one most important identity for solving any kind of problem with infinite product is converting it into a problem of infinite sums:
# \prod_{k=1}^{n}a_k = a_1 * a_2 * a_3 ... = e^{ln(a_1)} * e^{ln(a_2)} * e^{ln(a_3)} ... #
EMPHASIS:
# = exp[\sum_{k=1}^{n}ln(a_k)] #
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But, before we can do this, we must first deal with the #\frac{1}{n^2} in the equation and btw let's called the infinite product L:
# L = \lim_{n\to +\infty} \frac{1}{n^2}\prod_{k=1}^{n} (n^2 + k^2)^{\frac{1}{n}} #
# = \lim_{n\to +\infty} \frac{1}{n^2}\prod_{k=1}^{n} [n^2 (1+ \frac{k^2}{n^2})]^{\frac{1}{n}} #
# = \lim_{n\to +\infty} \frac{n^2}{n^2}\prod_{k=1}^{n} (1+ \frac{k^2}{n^2})^{\frac{1}{n}}
= \lim_{n\to +\infty} \prod_{k=1}^{n} (1+ \frac{k^2}{n^2})^{\frac{1}{n}} #
Now we can convert this into an infinite sum:
# L = \lim_{n\to +\infty} \prod_{k=1}^{n} (1+ \frac{k^2}{n^2})^{\frac{1}{n}}
= \lim_{n\to +\infty} exp[\sum_{k=1}^{n}ln((1+ \frac{k^2}{n^2})^{\frac{1}{n}})] #
apply logarithm properties:
# L = \lim_{n\to +\infty} exp[\sum_{k=1}^{n} \frac{1}{n} * ln(1+ \frac{k^2}{n^2})] #
And using limit properties:
# L = exp[\lim_{n\to +\infty} \sum_{k=1}^{n} \frac{1}{n} * ln(1+\frac{k^2}{n^2})] #
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Let's call the infinite sum S:
# S = \lim_{n\to +\infty} \sum_{k=1}^{n} \frac{1}{n} * ln(1+ \frac{k^2}{n^2}) #
And keep in mind that
# L = exp(S) #
Now let's solve your question by converting it from a RIEMANN SUM to a DEFINITE INTEGRAL:
Recall the definition of a Riemann sum is :
EMPHASIS:
# \int_{a}^{b}f(x)dx = \lim_{n\to +\infty}\sum_{k=1}^{n}f(a+k(\frac{b-a}{n})) * \frac{b-a}{n} #
Let
# \lim_{n\to +\infty}\sum_{k=1}^{n}f(a+k(\frac{b-a}{n})) * \frac{b-a}{n} = \lim_{n\to +\infty} \sum_{k=1}^{n} \frac{1}{n} * ln(1+ \frac{k^2}{n^2}) = S #
Now, let # f(x) = ln(1+x^2) and a = 0 #
# f(k(\frac{b}{n})) =ln(1+ \frac{k^2}{n^2}) #
Thus, b = 1 i.e.
# f(\frac{k}{n}) = ln(1+ \frac{k^2}{n^2}) #
Therefore,
# S = \lim_{n\to +\infty} \sum_{k=1}^{n} \frac{1}{n} * ln(1+ \frac{k^2}{n^2}) = \int_{0}^{1}ln(1+x^2)dx #
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Solve for # \int_{0}^{1}ln(1+x^2)dx #:
use integration by parts:
# \int uv dx = u\int v dx - \int (u' * \int vdx)dx #
Let # u = ln(1+x^2) and v = 1 #
Then, use chain rule and the derivative of natural logarithm to get # u' = 1/(1+x^2) * 2x = \frac{2x}{1+x^2}#
and use power rule to get: # \int 1dx = x #
# \int ln(1+x^2)dx = ln(1+x^2) * x - [ \int (\frac{2x}{1+x^2} * x)dx]#
# = ln(1+x^2) * x - [ \int \frac{2x^2}{1+x^2}dx]#
# = xln(1+x^2) - 2[ \int \frac{x^2}{1+x^2}dx]#
# = xln(1+x^2) - 2[ \int \frac{x^2+1 -1}{x^2 + 1}dx]# Use subtraction rule:
# = xln(1+x^2) - 2[ \int \frac{x^2+1}{x^2 + 1} - \int \frac{1}{x^2 + 1} dx]#
# = xln(1+x^2) - 2[ \int1 - \int \frac{1}{x^2 + 1} dx] #
Use power rule for the first integral and the second integral is the standard trigonometric function # arctan(x) # (the inverse of the tangent function)
# = xln(1+x^2) - 2[ x - arctan(x)] #
Thus, # \int ln(1+x^2)dx = xln(1+x^2) - 2x + 2 arctan(x) + C #
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Now solve for the definite integral:
# S = \int_{0}^{1}ln(1+x^2)dx #
we know that the anti-derivative is # F(x) = xln(1+x^2) - 2x + 2 arctan(x) + C#, Thus
# S = F(x) |_{x=0}^{x=1} = F(1) - F(0) #
#S = 1ln(1+1^2) - 2(1) + 2 arctan(1) - 0 + 0 - arctan(0) #
note that arctan(1) is 45° or # \frac{\pi}{4}# (recall the special right triangle with side lengths 1,1, #\sqrt{2}# and angles 45°,45°,90°) and also # arctan(0) = 0 #
Thus #S = ln(2) - 2 + 2 (\frac{\pi}{4}) = ln(2) - 2 + \frac{\pi}{2} #
or # \approx 0.263943507354...#
# L = exp[S]= exp[ln(2) - 2 + \frac{\pi}{2}] = e^{ln(2)} * e^{-2} * e^{\frac{\pi}{2}} #
# L = 2 * \frac{1}{e^2} * (e^{pi}) ^ {1/2}#
# L = \frac{2\sqrt{e^pi}}{e^2}#
Therefore the solution is # \lim_{n\to +\infty} \frac{1}{n^2}\prod_{k=1}^{n} (n^2 + k^2)^{\frac{1}{n}} = \frac{2\sqrt{e^pi}}{e^2}# or #\approx 1.302054638...#
