How to compute $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}}$ Trying to prove $\lim_{n \to \infty} \frac{1}{n}  \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}} = 2e^{-1/2}$
I tried using ratio-root criteria with $a_n= \frac{1}{n^n}  \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n}}$ and end up with $\frac{n^n}{(n+1)^{n+1}}\cdot 2(n+1)^2 \Pi_{k=1}^n\frac{\left(k^2\right)^{\frac{k}{n+1}} \left(1+\left(\frac{n+1}{k} \right)^2 \right)^{\frac{k}{n+1}}}{\left(k^2\right)^{\frac{k}{n}} \left(1+\left(\frac{n}{k} \right)^2 \right)^{\frac{k}{n}}}$ and can’t see how to simplify this to get the limit as the numerator and denominator seem to end up going to 1 and some kind of $e^{\Sigma \frac{1}{n}}$ but the $(n+1)^2$ seems to persist and send everything to infinity.
So I tried taking e^ natural log(I’ll denote them as log here, I don’t mind ln either) and got this sum in the exponent $-\log (n) + \Sigma_{k=1}^n \frac{k}{n^2}\log(n^2+k^2)$ comparison showed I could get a $\log 2$ and a $\log n$ But where is the $-1/2$ I need? My guess is that this approach won’t simplify further.
Am doing this for practice/review so would be interested in seeing solutions or hints  as I already spent too much time on this. My hunch is that I’m either overlooking something or there’s some typo in the problem.
This is also equivalent to $\lim_{n \to \infty} \frac{1}{n}  \left[(n^2 +1^2)^{\frac{1}{n^2}}(n^2+2^2)^{\frac{2}{n^2}} \cdots (n^2 + n^2)^{\frac{1}{n}} \right] = 2e^{-1/2}$
Try explaining this before calling this duplicates. I get something different:
$\frac{1}{n}  \left[(n^2 +1^2)^{1}(1+\frac{2}{n}^2)^{2} \cdots (n^2 + {n}^2)^{{n}} \right]^{\frac{1}{n^2}} = n^{\frac{1}{n}}\left[(1 +(\frac{1}{n})^2)^{\frac{1}{n^2}}(1+(\frac{2}{n})^2)^{\frac{2}{n^2}} \cdots (1 + (\frac{n}{n})^2)^{\frac{1}{n}} \right]$.
How do you get $\frac{1}{n}  \left[(n^2 +1^2)^{1}(1+\frac{2}{n}^2)^{2} \cdots (n^2 + {n}^2)^{{n}} \right]^{\frac{1}{n^2}} = (1 +(\frac{1}{n})^2)(1+(\frac{2}{n})^2)^{2} \cdots (1 + (\frac{n}{n})^2)^{n}$?

This result for the logarithm that you've got can be calculated further. We have
\begin{align} & -\log(n) + \sum_{k=1}^n \frac{k}{n^2} \log(k^2+n^2) = \\
&= -\log(n) + \sum_{k=1}^n \frac{k}{n^2} \Big(2\log n + \log(1+\frac{k^2}{n^2})\Big) = \\
&= -\log(n) + \frac{2\log n}{n^2}\frac{n(n+1)}{2} + \frac{1}{n} \sum_{k=1}^n\frac{k}{n}\log(1+\frac{k^2}{n^2}) = \\
&= \frac{\log n}{n} + \frac{1}{n} \sum_{k=1}^n \frac{k}{n}\log(1+\frac{k^2}{n^2}) 
\end{align}
In the limit
$$ \lim_{n\to\infty} (\dots) = 0 + \int_0^1 x\log(1+x^2) dx = \frac12(2\log 2 -1)$$
