# Solving this using riemann integral?

May 21, 2018

$\setminus \frac{2 \setminus \sqrt{{e}^{\pi}}}{{e}^{2}}$ or $\setminus \approx 1.302054638 \ldots$

#### Explanation:

The number one most important identity for solving any kind of problem with infinite product is converting it into a problem of infinite sums:

$\setminus {\prod}_{k = 1}^{n} {a}_{k} = {a}_{1} \cdot {a}_{2} \cdot {a}_{3} \ldots = {e}^{\ln \left({a}_{1}\right)} \cdot {e}^{\ln \left({a}_{2}\right)} \cdot {e}^{\ln \left({a}_{3}\right)} \ldots$

EMPHASIS:

$= \exp \left[\setminus {\sum}_{k = 1}^{n} \ln \left({a}_{k}\right)\right]$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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:

$L = \setminus {\lim}_{n \setminus \to + \setminus \infty} \setminus \frac{1}{{n}^{2}} \setminus {\prod}_{k = 1}^{n} {\left({n}^{2} + {k}^{2}\right)}^{\setminus \frac{1}{n}}$

$= \setminus {\lim}_{n \setminus \to + \setminus \infty} \setminus \frac{1}{{n}^{2}} \setminus {\prod}_{k = 1}^{n} {\left[{n}^{2} \left(1 + \setminus \frac{{k}^{2}}{{n}^{2}}\right)\right]}^{\setminus \frac{1}{n}}$

 = \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:

$L = \setminus {\lim}_{n \setminus \to + \setminus \infty} \exp \left[\setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{n} \cdot \ln \left(1 + \setminus \frac{{k}^{2}}{{n}^{2}}\right)\right]$

And using limit properties:

$L = \exp \left[\setminus {\lim}_{n \setminus \to + \setminus \infty} \setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{n} \cdot \ln \left(1 + \setminus \frac{{k}^{2}}{{n}^{2}}\right)\right]$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let's call the infinite sum S:
$S = \setminus {\lim}_{n \setminus \to + \setminus \infty} \setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{n} \cdot \ln \left(1 + \setminus \frac{{k}^{2}}{{n}^{2}}\right)$

And keep in mind that
$L = \exp \left(S\right)$

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:
$\setminus {\int}_{a}^{b} f \left(x\right) \mathrm{dx} = \setminus {\lim}_{n \setminus \to + \setminus \infty} \setminus {\sum}_{k = 1}^{n} f \left(a + k \left(\setminus \frac{b - a}{n}\right)\right) \cdot \setminus \frac{b - a}{n}$

Let
$\setminus {\lim}_{n \setminus \to + \setminus \infty} \setminus {\sum}_{k = 1}^{n} f \left(a + k \left(\setminus \frac{b - a}{n}\right)\right) \cdot \setminus \frac{b - a}{n} = \setminus {\lim}_{n \setminus \to + \setminus \infty} \setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{n} \cdot \ln \left(1 + \setminus \frac{{k}^{2}}{{n}^{2}}\right) = S$

Now, let $f \left(x\right) = \ln \left(1 + {x}^{2}\right) \mathmr{and} a = 0$
$f \left(k \left(\setminus \frac{b}{n}\right)\right) = \ln \left(1 + \setminus \frac{{k}^{2}}{{n}^{2}}\right)$
Thus, b = 1 i.e.
$f \left(\setminus \frac{k}{n}\right) = \ln \left(1 + \setminus \frac{{k}^{2}}{{n}^{2}}\right)$

Therefore,
$S = \setminus {\lim}_{n \setminus \to + \setminus \infty} \setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{n} \cdot \ln \left(1 + \setminus \frac{{k}^{2}}{{n}^{2}}\right) = \setminus {\int}_{0}^{1} \ln \left(1 + {x}^{2}\right) \mathrm{dx}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Solve for $\setminus {\int}_{0}^{1} \ln \left(1 + {x}^{2}\right) \mathrm{dx}$:

use integration by parts:
$\setminus \int u v \mathrm{dx} = u \setminus \int v \mathrm{dx} - \setminus \int \left(u ' \cdot \setminus \int v \mathrm{dx}\right) \mathrm{dx}$

Let $u = \ln \left(1 + {x}^{2}\right) \mathmr{and} v = 1$
Then, use chain rule and the derivative of natural logarithm to get $u ' = \frac{1}{1 + {x}^{2}} \cdot 2 x = \setminus \frac{2 x}{1 + {x}^{2}}$
and use power rule to get: $\setminus \int 1 \mathrm{dx} = x$

$\setminus \int \ln \left(1 + {x}^{2}\right) \mathrm{dx} = \ln \left(1 + {x}^{2}\right) \cdot x - \left[\setminus \int \left(\setminus \frac{2 x}{1 + {x}^{2}} \cdot x\right) \mathrm{dx}\right]$

$= \ln \left(1 + {x}^{2}\right) \cdot x - \left[\setminus \int \setminus \frac{2 {x}^{2}}{1 + {x}^{2}} \mathrm{dx}\right]$

$= x \ln \left(1 + {x}^{2}\right) - 2 \left[\setminus \int \setminus \frac{{x}^{2}}{1 + {x}^{2}} \mathrm{dx}\right]$

$= x \ln \left(1 + {x}^{2}\right) - 2 \left[\setminus \int \setminus \frac{{x}^{2} + 1 - 1}{{x}^{2} + 1} \mathrm{dx}\right]$ Use subtraction rule:

$= x \ln \left(1 + {x}^{2}\right) - 2 \left[\setminus \int \setminus \frac{{x}^{2} + 1}{{x}^{2} + 1} - \setminus \int \setminus \frac{1}{{x}^{2} + 1} \mathrm{dx}\right]$

$= x \ln \left(1 + {x}^{2}\right) - 2 \left[\setminus \int 1 - \setminus \int \setminus \frac{1}{{x}^{2} + 1} \mathrm{dx}\right]$
Use power rule for the first integral and the second integral is the standard trigonometric function $\arctan \left(x\right)$ (the inverse of the tangent function)

$= x \ln \left(1 + {x}^{2}\right) - 2 \left[x - \arctan \left(x\right)\right]$
Thus, $\setminus \int \ln \left(1 + {x}^{2}\right) \mathrm{dx} = x \ln \left(1 + {x}^{2}\right) - 2 x + 2 \arctan \left(x\right) + C$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now solve for the definite integral:
$S = \setminus {\int}_{0}^{1} \ln \left(1 + {x}^{2}\right) \mathrm{dx}$

we know that the anti-derivative is $F \left(x\right) = x \ln \left(1 + {x}^{2}\right) - 2 x + 2 \arctan \left(x\right) + C$, Thus

$S = F \left(x\right) {|}_{x = 0}^{x = 1} = F \left(1\right) - F \left(0\right)$

$S = 1 \ln \left(1 + {1}^{2}\right) - 2 \left(1\right) + 2 \arctan \left(1\right) - 0 + 0 - \arctan \left(0\right)$

note that arctan(1) is 45­° or $\setminus \frac{\setminus \pi}{4}$ (recall the special right triangle with side lengths 1,1, $\setminus \sqrt{2}$ and angles 45°,45°,90°) and also $\arctan \left(0\right) = 0$

Thus $S = \ln \left(2\right) - 2 + 2 \left(\setminus \frac{\setminus \pi}{4}\right) = \ln \left(2\right) - 2 + \setminus \frac{\setminus \pi}{2}$
or $\setminus \approx 0.263943507354 \ldots$

$L = \exp \left[S\right] = \exp \left[\ln \left(2\right) - 2 + \setminus \frac{\setminus \pi}{2}\right] = {e}^{\ln \left(2\right)} \cdot {e}^{- 2} \cdot {e}^{\setminus \frac{\setminus \pi}{2}}$

$L = 2 \cdot \setminus \frac{1}{{e}^{2}} \cdot {\left({e}^{\pi}\right)}^{\frac{1}{2}}$

$L = \setminus \frac{2 \setminus \sqrt{{e}^{\pi}}}{{e}^{2}}$

Therefore the solution is $\setminus {\lim}_{n \setminus \to + \setminus \infty} \setminus \frac{1}{{n}^{2}} \setminus {\prod}_{k = 1}^{n} {\left({n}^{2} + {k}^{2}\right)}^{\setminus \frac{1}{n}} = \setminus \frac{2 \setminus \sqrt{{e}^{\pi}}}{{e}^{2}}$ or $\setminus \approx 1.302054638 \ldots$