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. 
 A: For any $n$ this is a Riemann sum of the integral $$ \int_0^1\log\left((1+\frac{x}{n})^n\right)dx$$ and its error is less than $1/n$.  Now the integrand converges to $x$ for $n\to\infty$ so these sums converge to $\tfrac{1}{2}$.
A: Your answer can't be right
We have, if $x_k>0$, that $$\prod_{k=1}^n \left(1+x_k\right) \geq 1+x_1+x_2+\cdots+x_n$$
So $$\prod_{k=1}^n \left(1+\frac{k}{n^2}\right) \geq 1+\frac{n(n+1)}{2n^2} = 1+\frac{1}{2}+\frac{1}{2n}$$
So , as $n\to\infty$, this is at least $\frac{3}{2}$, so your limit, if it exists, must be greater than $\log\left(\frac32\right)>0$.
Correct Answer
Using that, for $0<x<1$, $\log(1+x) = \sum_{i=1}^\infty (-1)^{i-1}\frac{x^i}{i}$, we can see that:
$$\sum_{k=1}^n \log\left(1+\frac{k}{n^2}\right) = \sum_{i=1}^\infty \frac{(-1)^{i-1}}{in^{2i}}\sum_{k=1}^{n} k^i$$
But $\sum_{k=1}^n k^i \leq n^{i+1}$, so $$\frac{1}{n^{2i}}\sum_1^n k^i < \frac{1}{n^{i-1}}$$ and therefore,  $$\left|\sum_{i=2}^\infty  \frac{(-1)^{i-1}}{in^{2i}}\sum_{k=1}^{n} k^i\right|<\sum_{i=2}^\infty \frac {1}{n^{i-1}} = \frac{1}{n-1}\to 0$$
So the only term that matters is $i=1$ as $n\to\infty$. 
The first term is $\frac{1}{n^2}\sum_{k=1}^n k = \frac{n^2+n}{2n^2} \to \frac12$.
A: Another way of writing your sum is
$$
\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)=\log\left(\frac{(n^2+n)!}{n^2!}\right)-2n\log(n)
$$
We can use Stirling's Formula to get
$$
\log((n^2+n)!)=(n^2+n)\log(n^2+n)-(n^2+n)+\frac12\log(2\pi(n^2+n))+O\left(\frac1{n^2}\right)
$$
and
$$
\begin{align}
\log(n^2!)
&=n^2\log(n^2)-n^2+\frac12\log(2\pi n^2)+O\left(\frac1{n^2}\right)\\
&=(n^2+n)\log(n^2)-n\log(n^2)-n^2+\frac12\log(2\pi n^2)+O\left(\frac1{n^2}\right)
\end{align}
$$
Subtracting yields
$$
\begin{align}
\log\left(\frac{(n^2+n)!}{n^2!}\right)
&=\left(n^2+n+\frac12\right)\log\left(1+\frac1n\right)+2n\log(n)-n+O\left(\frac1{n^2}\right)\\
&=\left(n^2+n+\frac12\right)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}+O\left(\frac1{n^4}\right)\right)\\
&+2n\log(n)-n+O\left(\frac1{n^2}\right)\\
&=2n\log(n)+\frac12+\frac1{3n}+O\left(\frac1{n^2}\right)
\end{align}
$$
Therefore, we get
$$
\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)=\frac12+\frac1{3n}+O\left(\frac1{n^2}\right)
$$
and the limit
$$
\lim_{n\to\infty}\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)=\frac12
$$
A: 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 ${ 1\over 2}$.
A: $$S=\lim_{n\to\infty}\sum_{k=1}^n\ln(1+k/n^2)$$
$$S=\lim_{n\to\infty}\ln\left(\prod_{k=1}^n(1+k/n^2)\right)=^*\ln(\sqrt e)=\frac12\ne0$$
__
$^*\displaystyle \lim_{n\to\infty} (1/n^2)^n (1+n^2)_n = \sqrt e$ from here
A: $ \lim_{n\to \infty} \frac{m}{n} \cdot \sum_{k=1}^n \ln (1+\frac{k}{mn}) = m \cdot\int_0^1 \ln(1+\frac{x}{m}) = m(m+1)\ln(1+\frac{1}{m})-m$.
This last expression approaches $\frac{1}{2}$ as $m\to \infty$.
This is not quite a proof, but we can make it into one by carefully bounding the difference between the sum and the integral.
