Ignoring $\frac1n$ in Riemann Sum I have $$\ln y =\lim_{n\to\infty}{1\over2n}\sum^n_{k=1}\ln\left(1+\frac{2k-1}{2n}\right)\\=\frac12\lim_{n\to\infty}{1\over n}\sum^n_{k=1}\ln\left(1+\frac{k-1/2}{n}\right)$$
Can I just ignore the $-{1\over2n}$ part (inside $\ln$) and evaluate this as $$\ln y =\frac12\int_0^1\ln(1+x)\,dx$$$$\implies y=\frac2{\sqrt e}$$
Why?/ Why not?
 A: As regards your first comment, you are not actually ignoring the $-\frac{1}{2n}$. According to the definition of Riemann integral, you have
$$\int_a^b f(x)\,dx = \lim_{n \to +\infty}\frac{b-a}{n}\sum_{i=1}^n f(\xi_i),$$
where the $\xi_i$'s are $n$ points in the sub-intervals of the equipartition you are considering (in particular, $\xi_i$ lies in the $i$-th sub-interval). In your case $a=0, b=1$, $f(x) = \log(1+x)$ and $\xi_i = \frac{i-1/2}{n}$. Notice that 
$$\frac{i-1}{n} \le \frac{i-1/2}{n} \le \frac{i}{n} $$
so $\xi_i$ is a point in the $i$'th sub-interval of the partition, which fits with the definition above.
A: Well, since 
$$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right) = \int_{0}^{1}\log(1+x)\,dx = -1+2\log 2 \tag{1}$$
we also have
$$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k-\frac{1}{2}}{n}\right) = \int_{0}^{1}\log(1+x)\,dx = -1+2\log 2 \tag{2}$$
since the difference between the LHS of $(2)$ and the LHS of $(1)$ is
$$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(1-\frac{1}{2n+2k}\right)=\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\Theta\left(\frac{1}{n}\right)=\lim_{n\to +\infty}\Theta\left(\frac{1}{n}\right)=0.\tag{3}$$
