Infinite product

$Let:\quad f\left( x \right) :=1+\frac { 1 }{ { x }^{ 2 } } \qquad { t }_{ i }=1+\frac { i }{ n } \\ r=\lim _{ n\rightarrow \infty }{ \sqrt [ n ]{ 1+\frac { { n }^{ 2 } }{ \left( { n }+1 \right) ^{ 2 } } } \cdot \sqrt [ n ]{ 1+\frac { { n }^{ 2 } }{ \left( { n }+2 \right) ^{ 2 } } } \cdot ...\cdot \sqrt [ n ]{ 1+\frac { { n }^{ 2 } }{ \left( { n }+{ n }^{ 4 } \right) ^{ 2 } } } } =\\ =\lim _{ n\rightarrow \infty }{ \prod _{ i=1 }^{ { n }^{ 4 } }{ \left( 1+\frac { { n }^{ 2 } }{ \left( { n }+i \right) ^{ 2 } } \right) ^{ \frac { 1 }{ n } } } } \Leftrightarrow \ln { r } =\lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ { n }^{ 4 } }{ \frac { 1 }{ n } \cdot \ln { \left( 1+\frac { { n }^{ 2 } }{ \left( n\left( 1+\frac { i }{ n } \right) \right) ^{ 2 } } \right) } } } =\\ =\lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ { n }^{ 4 } }{ \frac { 1 }{ n } \cdot \ln { \left( 1+\frac { 1 }{ \left( 1+\frac { i }{ n } \right) ^{ 2 } } \right) } } } =\lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ { n }^{ 4 } }{ \frac { 1 }{ n } \cdot \ln { \left( 1+\frac { 1 }{ \left( { t }_{ i } \right) ^{ 2 } } \right) } } } =\lim _{ n\rightarrow \infty }{ \sum _{ i=1 }^{ { n }^{ 4 } }{ \frac { 1 }{ n } \cdot \ln { f\left( { t }_{ i } \right) } } } =\\ =\lim _{ n\rightarrow \infty }{ \int _{ 1 }^{ { n }^{ 3 }+1 }{ \ln { \left( 1+{ x }^{ -2 } \right) } } dx } \qquad now\quad let:\begin{cases} u=\ln { \left( 1+{ x }^{ -2 } \right) } \Leftrightarrow du=\frac { -2{ x }^{ -3 } }{ 1+{ x }^{ -2 } } dx \\ dv=dx\Leftrightarrow v=x \end{cases}\\ integrating\quad by\quad parts,\quad it's:\quad \int { udv } =uv-\int { vdu } =x\ln { \left( 1+{ x }^{ -2 } \right) } +2\int { \frac { x\cdot { x }^{ -3 } }{ 1+{ x }^{ -2 } } dx } =\\ =x\ln { \left( 1+{ x }^{ -2 } \right) } +2\int { \frac { 1 }{ { x }^{ 2 }+1 } dx } =x\ln { \left( 1+{ x }^{ -2 } \right) } +2\arctan { x } +C\\ So:\quad \ln { r } =\lim _{ n\rightarrow \infty }{ \int _{ 1 }^{ { n }^{ 3 }+1 }{ \ln { \left( 1+{ x }^{ -2 } \right) } } dx } =\overbrace { x\ln { \left( 1+{ x }^{ -2 } \right) } }^{ \infty \cdot 0 } +2\overbrace { \arctan { \infty } }^{ \frac { \pi }{ 2 } } -\left( \ln { 2 } +2\overbrace { \arctan { 1 } }^{ \frac { \pi }{ 4 } } \right) \\ \lim _{ x\rightarrow +\infty }{ x\ln { \left( 1+{ x }^{ -2 } \right) } } =\lim _{ x\rightarrow +\infty }{ \frac { \ln { \left( 1+{ x }^{ -2 } \right) } }{ { x }^{ -1 } } } =\overbrace { \lim _{ x\rightarrow +\infty }{ \frac { \frac { -2x^{ -3 } }{ 1+{ x }^{ -2 } } }{ -{ x }^{ -2 } } } }^{ by\quad L'Hopital\quad Rule } =\lim _{ x\rightarrow +\infty }{ \frac { 2x^{ -3 } }{ { x }^{ -2 }+{ x }^{ -4 } } = } \lim _{ x\rightarrow +\infty }{ \frac { 2x }{ { x }^{ 2 }+1 } = } 0\\ \\ \therefore \ln { r } =\pi -\ln { 2 } -\frac { \pi }{ 2 } \Leftrightarrow \boxed { r=\frac { \sqrt { { e }^{ \pi } } }{ 2 } \cong 2.405 }$

Note : I've added this in response to some questions in the comments in Oleg's post. I'm going to show one way to justify using an improper Riemann integral to find the answer by way of the Lebesgue Dominated Convergence theorem. At this point, the solution would follow Oleg's almost exactly, so I will just reference his solution for that part.

By applying the natural logarithm and using continuity, $\ln r = \lim_{n\rightarrow\infty} \sum_{i=1}^{n^4} \frac{1}{n} \ln\left(1 + \frac{n^2}{(n+i)^2}\right) = \lim_{n\rightarrow\infty} \sum_{i=1}^{n^4} \frac{1}{n} \ln\left(1 + \frac{1}{\left(1+\tfrac{i}{n}\right)^2}\right).$

To investigate this limit, we define $f(x) = \ln\left(1 + \frac{1}{(1+x)^2}\right), \qquad f_n(x) = \begin{cases} f\left(\frac{1}{n}\lceil{nx}\rceil\right) & 0 < x \leq n^3 \\ 0 & \mbox{otherwise}\end{cases}$ and note that $\begin{aligned}\int\limits_{(0,\infty)} f_n(x)\,dx &= \int\limits_{(0, n^3]} f\left(\tfrac{1}{n}\lceil{nx}\rceil\right)\,dx \\ &= \sum_{i=1}^{n^4} \int\limits_{\left(\tfrac{i-1}{n}, \tfrac{i}{n}\right]} f\left(\tfrac{1}{n}\lceil{nx}\rceil\right)\,dx \\ &= \sum_{i=1}^{n^4} \int\limits_{\left(\tfrac{i-1}{n}, \tfrac{i}{n}\right]} f\left(\tfrac{i}{n}\right)\,dx \\ &= \sum_{i=1}^{n^4} f\left(\tfrac{i}{n}\right)\left(\frac{i}{n} - \frac{i-1}{n}\right) \\ &= \sum_{i=1}^{n^4} \frac{1}{n} \ln\left(1 + \frac{1}{\left(1+\tfrac{i}{n}\right)^2}\right),\end{aligned}$ so $\ln r = \lim_{n\rightarrow\infty} \int\limits_{(0,\infty)} f_n(x)\,dx.$

Then, since $f$ is non-negative, non-increasing, and right-continuous on $x>0$ , we can see that

• $\displaystyle\lim_{n\rightarrow\infty} f_n(x) = f(x)$ for all $x > 0$ . [by right-continuity]
• $0 \leq f_n(x) \leq f(x)$ for all $n$ and $x > 0$ . [the inequalities follow by non-negativity and non-increasing, respectively]

Therefore, by the Dominated Convergence theorem, we can conclude that IF $\int_{(0,\infty)} f(x)\, dx < \infty$ , then $\ln r = \int_{(0,\infty)} f(x)\,dx,$ and since $f(x)$ is positive, we may evaluate this Lebesgue integral as the improper Riemann integral $\int_0^\infty \ln\left(1 + \frac{1}{(1+x)^2}\right) \,dx = \int_1^\infty \ln\left(1 + \frac{1}{x^2}\right)\,dx = \lim_{N\rightarrow\infty} \int_1^N \ln\left(1 + \frac{1}{x^2}\right)\,dx$

Since this improper integral equals $\frac{\pi}{2} - \ln 2$ [see Oleg's solution], which is finite, we may conclude by the above remarks that $\ln r = \frac{\pi}{2} - \ln 2$