Using the exp-log technique,
[tex]\displaystyle \lim_{n\to\infty} \frac{\left(1 \times 2^2 \times 3^3 \times \cdots \times n^n\right)^{\frac1{n^2}}}{n^{\frac{n+1}{2n}}}[/tex]
[tex]\displaystyle = \exp\left(\lim_{n\to\infty} \frac{\ln\left(\prod\limits_{k=1}^n k^k\right)}{n^2} - \frac{n+1}{2n} \ln(n)\right)\right)[/tex]
[tex]\displaystyle = \exp\left(\lim_{n\to\infty} \frac{\sum\limits_{k=1}^n k\ln(k)}{n^2} - \frac{n+1}{2n} \ln(n)\right)\right)[/tex]
We have ln(k) = ln(k/n) + ln(n), and so we can rewrite the limand as
[tex]\displaystyle = \exp\left(\lim_{n\to\infty} \left(\frac1n \sum\limits_{k=1}^n \frac kn \ln\left(\frac kn\right) + \frac{\ln(n)}{n^2} \sum_{k=1}^n 1 - \frac{n+1}{2n} \ln(n)\right)\right)[/tex]
Then the first sum converges to a definite integral,
[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \frac kn \ln\left(\frac kn\right) = \int_0^1 x \ln(x) \, dx = -\frac14[/tex]
while the remaining terms vanish since
[tex]\displaystyle \frac{\ln(n)}{n^2} \sum_{k=1}^n 1 = \frac{\ln(n)}{n^2} \times \frac{n(n+1)}2 = \frac{n+1}{2n}\ln(n)[/tex]
So the limit is [tex]\boxed{e^{-1/4}}[/tex].
Since
[tex]\dfrac1{x(x+1)} = \dfrac1x - \dfrac1{x+1}[/tex]
it's easy to show that the integral reduces to
[tex]\displaystyle \int_k^{k+1} \frac{k+1}{x(x+1)} \, dx = (k+1) \ln\left(\dfrac{(k+1)^2}{k(k+2)}\right)[/tex]
so we can write the sum as
[tex]I = \displaystyle \sum_{k=1}^{98} \int_k^{k+1} \frac{k+1}{x(x+1)} \, dx = \sum_{k=2}^{99} k \ln\left(\frac{k^2}{k^2-1}\right)[/tex]
We have k²/(k² - 1) > 1 for all k, so that ln(k²/(k² - 1)) > ln(1) = 0. We can see the first 3 terms of the sum already exceed 1 > 49/50, so (D) is true.
Numerical computation of the sum suggests I < ln(99), but I have yet to come up with an analytical solution for this bound...
