Limit using Stirling's formula Consider the limit
$$\lim_{n \to \infty} \left[n(1-2\log2) + \sum_{k=1}^{n} \log\left(1+\frac {k}{n}\right)\right] = \frac {1}{2} \log 2.$$
This can be shown by using Stirling's formula for $n!$. My question is if this is the only way or there is also an elementary solution.
 A: This is the case $f:x\mapsto\log(1+x)$ of the more general result that, for every $C^1$ function $f$ on $[0,1]$,

$$\lim_{n \to \infty} \sum_{k=1}^{n} f\left(\frac {k}{n}\right)-n\int_0^1f(x)\mathrm dx = \frac {1}{2}(f(1)-f(0)).$$

To show this, one can use Riemann sums and a first-order Taylor expansion on $f$. 
More in details, one starts from the identity $$f\left(\frac {k}{n}\right)=n\int_{(k-1)/n}^{k/n}f(x)\mathrm dx+R_{k,n},\qquad R_{k,n}=n\int_{(k-1)/n}^{k/n}\left(f\left(\frac {k}{n}\right)-f(x)\right)\mathrm dx.$$ The integrand in $R_{k,n}$ is $$f\left(\frac {k}{n}\right)-f(x)=f'\left(\frac{k}n\right)\cdot\left(\frac {k}{n}-x\right)+o\left(\frac1n\right),$$ hence $$R_{k,n}=nf'\left(\frac{k}n\right)\int_{(k-1)/n}^{k/n}\left(\frac {k}{n}-x\right)\mathrm dx+o\left(\frac1{n}\right)=f'\left(\frac{k}n\right)\frac1{2n}+o\left(\frac1{n}\right).$$ Summing these from $k=1$ to $k=n$ yields $$\sum_{k=1}^{n} f\left(\frac {k}{n}\right)=n\int_0^1f(x)\mathrm dx +\frac12\frac1n\sum_{k=1}^{n}f'\left(\frac{k}n\right)+o(1),$$ and it remains to identify the sum on the RHS as a Riemann sum of $f'$ to deduce that $$\frac1n\sum_{k=1}^{n}f'\left(\frac{k}n\right)=\int_0^1f'(x)\mathrm dx+o(1)=f(1)-f(0)+o(1),$$ hence the identity above.
