<s>[INST] What is the limit of the following sum $$\lim_{n\to\infty}\sum_{k=1}^n \ln\Big(1+\frac{k}{n^2}\Big)$$
According to me, the answer is $0$. I'm curious as to what answers might others come up with, as well as the method of reasoning. 
 [/INST] Rather boring answer:
We have $x-{x^2 \over 2} \le \log(1+x) \le x$ for all $x >0$.
We have
$\sum_{k=1}^n  \log(1+ { k \over n^2}) \le \sum_{k=1}^n { k \over n^2} = {1 \over n^2} (1+2 + \cdots + n) = {1 \over n^2} {1 \over 2} n (n+1)$.
We also have $\sum_{k=1}^n \log(1+ { k \over n^2}) \ge  \sum_{k=1}^n ( { k \over n^2} -{ 1\over 2} ({ k \over n^2})^2 ) = {1 \over n^2} {1 \over 2} n (n+1) - { 1\over 2} {1 \over n^4} (1^2+2^2+\cdots + n^2)$.
Since $1^2+2^2+\cdots + n^2 = {1 \over 3} n^3 + {1 \over 2} n^2 +{1 \over 6} n$, we 
 see that the limit of the sum is ${ 1\over 2}$.
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