$\lim_{n \to \infty} \sqrt[n^2]{\frac{(n+1)^{n+1}(n+2)^{n+2} \cdot (...) \cdot (2n)^{2n}}{n^{n+1}n^{n+2} \cdot (...) \cdot n^{2n}}}$? How to find the limit?
$$\lim_{n \to \infty} \sqrt[n^2]{\frac{(n+1)^{n+1}(n+2)^{n+2} \cdot (...) \cdot (2n)^{2n}}{n^{n+1}n^{n+2} \cdot (...) \cdot n^{2n}}}$$
The question is asked that way, so I suppose the limit exists. It's the last exercise from a chapter of my book where the previous ones I did using Riemann sum. Because of that I thought about log function to get change multiplications into sums. However, in the numerator, the basis of every exponential function is different.
Searching this site I found that Stirlings formula is being used in similar cases, however I don't know how to apply it nor if it's even the thing to do.
 A: Let's say
$$L=\lim_{n \to \infty} \sqrt[n^2]{\frac{(n+1)^{n+1}(n+2)^{n+2} \cdot \ldots \cdot (2n)^{2n}}{n^{n+1}n^{n+2} \cdot \ldots \cdot n^{2n}}}.$$
Then
\begin{align*}
\log(L)&=\lim_{n\to\infty}{\frac{1}{n^2}\log\left(\frac{(n+1)^{n+1}(n+2)^{n+2} \cdot \ldots \cdot (2n)^{2n}}{n^{n+1}n^{n+2} \cdot \ldots \cdot n^{2n}}\right)}\\
&=\lim_{n\to\infty}{\frac{1}{n^2}\log\left(\left(\frac{n+1}{n}\right)^{n+1}\cdot \left(\frac{n+2}{n}\right)^{n+2}\cdot\ldots\cdot\left(\frac{n+n}{n}\right)^{n+n}\right)}\\
&=\lim_{n\to\infty}{\frac{1}{n^2}\sum_{k=1}^{n}{(n+k)\log\left(\frac{n+k}{n}\right)}}\\
&=\lim_{n\to\infty}{\frac{1}{n}\sum_{k=1}^{n}{\left(1+\frac{k}{n}\right)\log\left(1+\frac{k}{n}\right)}}\\
\end{align*}
and this looks like a Riemann sum. Can you follow from here?
A: Just for your curiosity
Trying to go beyond the limit (using special functions)
We have
$$a_n=\Bigg[{\frac{\prod_{k=1}^n (n+k)^{(n+k)}}{\prod_{k=1}^n n^{(n+k)}}}\Bigg]^{\frac 1 {n^2}}$$ Taking logarithms
$$n^2\,\log(a_n)=\sum_{k=1}^n (n+k)\log(n+k)-\sum_{k=1}^n (n+k)\log(n)=\sum_{k=1}^n (n+k)\log \left(1+\frac{k}{n}\right)$$
$$\sum_{k=1}^n (n+k)\log \left(1+\frac{k}{n}\right)=-\zeta ^{(1,0)}(-1,n+1)+\zeta ^{(1,0)}(-1,2 n+1)-\frac{1}{2} n (3 n+1) \log (n)$$where appear the derivative of the $\zeta$ function with respect to the first argument.
Now, for large $n$
$$n^2\,\log(a_n)=n^2 \left(2 \log (2)-\frac{3}{4}\right)+n \log (2)+\frac{\log (2)}{12}-\frac{1}{960
   n^2}+O\left(\frac{1}{n^4}\right)$$
$$\log(a_n)=\left(2 \log (2)-\frac{3}{4}\right)+\frac{\log (2)}{n}+\frac{\log (2)}{12
   n^2}-\frac{1}{960 n^4}+O\left(\frac{1}{n^6}\right)$$
$$a_n=e^{\log(a_n)}=\frac{4}{e^{3/4}}\Bigg[1+ \frac{\log (2)}{n}+\frac{\log (2) (1+6 \log (2))}{12 n^2}\Bigg]+O\left(\frac{1}{n^3}\right)$$
For $n=10$, this truncated series gives
$$a_{10}=\frac{30 (40+4 \log (2))+\log (2) (1+6 \log (2))}{300 e^{3/4}}=2.0260644$$ while the exact value is $2.0262496$.
