Try to prove $\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$ $$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
Try to prove $$\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$$
I try to decompose $\ln 2$ as $$\ln(2n)-\ln(n)=\ln\left(1+\frac{1}{2n-1}\right)+\dots+\ln\left(1+\frac{1}{n}\right)\;,$$ but I can't continue, is that right?
 A: Letting $f(x)=1/x$ we have
$$
\int_1^2 f(x) \; dx = \log_e 2
$$
and
$$
\frac 1 n \left( f\left(1+\frac 1 n\right) + f\left(1+\frac 2 n\right) + f\left(1+\frac 3 n\right) + \cdots + f\left(1+\frac n n\right) \right) \to \int_1^2 f(x) \; dx\text{ as }n\to\infty,
$$
so
$$
\frac 1 n \left( \frac{n}{n+1} + \frac{n}{n+2}+\frac{n}{n+3} + \cdots + \frac{n}{n+n} \right) \to \log_e 2 \text{ as } n\to \infty. \tag{1}
$$
Since $f$ is a decreasing function, this is a lower Riemann sum, so it's approaching the integral from below.  The difference
$$
\log_e 2 - \{\text{the sum in (1)}\}
$$
is positive and approaches $0$.  That difference is the sum of the areas of $n$ regions below the curve and above the tops of the rectangles that you draw when you illustrate the Riemann sum.  Each such region is almost a triangle.  Its base has length $1/n$.  It has a vertical side that is a straight line.  Its hypotenuse is a curve that is nearly a straight line.  The sum of the heights of those almost-triangles is $1/2$.  So the sum of $1/2\times\text{base}\times\text{height}$ is $1/2\times1/n\times1/2$.  Multiply it by $n$ to get $1/4$.
But they're not exactly triangles, since the hypotenuse is a curve.  It approaches a straight line as $n\to\infty$.  The remaining problem is to deal with this present paragraph.
A: Let's cheat and use one of Euler's many results:
$$\sum_{i=1}^{n} \frac{1}{i} = \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$$
Note that:
$$A_n + \sum_{i=1}^{n} \frac{1}{i} = \sum_{i=1}^{2n} \frac{1}{i}$$
Substituting Euler's result for both summations, we get:
$$A_n + \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right) = \ln 2n + \gamma + \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$$
Rearranging, and using $\ln 2n = \ln 2 + \ln n$, we get
$$A_n = \ln 2 - \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$$
Thus the requested limit becomes
$$\lim_{n \to \infty} n (\ln 2 - A_n) = \lim_{n \to \infty} n \left(\frac{1}{4n} - O\left(\frac{1}{n^2}\right)\right) = \frac{1}{4}$$
A: Trying to follow the idea of the OP, write $$ \log 2 = \log (2n+2) - \log (n+1) = \log \left( 1+ \frac{1}{2n+1} \right) + \log \left( 1+ \frac{1}{2n} \right) + \cdots + \log \left( 1+ \frac{1}{n+1} \right) $$
so then $$ n( \log 2 - A_n) = n\log \left(1+ \frac{1}{2n+1} \right) + n\sum_{k=1}^n \left( \log \left( 1+ \frac{1}{n+k} \right) - \frac{1}{n+k} \right) .$$
Since near $x=0$ we have $\displaystyle \log(1+x) = x - \frac{x^2}{2} + \mathcal{O}(x^3) ,$ the first term tends to $1/2$ and the summand is $ \displaystyle \frac{-1}{2 (n+k)^2 } + \mathcal{O}(1/n^3) .$ Thus, $$n\sum_{k=1}^n \left( \log \left( 1+ \frac{1}{n+k} \right) - \frac{1}{n+k} \right) = \frac{-1}{2} \cdot \frac{1}{n} \left( \sum_{k=1}^n \frac{1}{\left(1+ \frac{k}{n} \right)^2}\right) + \mathcal{O}(1/n) $$
$$ \to \frac{-1}{2} \int^1_0 \frac{1}{(1+x)^2} dx= -\frac{1}{4}. $$
Thus, $$ n(\log 2 - A_n) \to \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.$$
A: Note that
$$
\log2-A_n=\int_n^{2n}\frac{\mathrm dx}x-\sum_{k=1}^n\frac1{n+k}=\sum_{k=1}^n\int_0^1\frac{1-x}{(n+k-1+x)(n+k)}\mathrm dx.
$$
To get an upper bound, use $n+k-1+x\geqslant n+k-1$, hence
$$
\log2-A_n\leqslant\sum_{k=1}^n\frac{1}{(n+k-1)(n+k)}\int_0^1(1-x)\mathrm dx.
$$
The integral is $\frac12$ and the sum is
$$
\sum_{k=1}^n\left(\frac{1}{n+k-1}-\frac1{n+k}\right)=\frac1n-\frac1{2n}=\frac1{2n},
$$
hence
$$
\log2-A_n\leqslant\frac1{4n}.
$$
To get a lower bound, use $n+k-1+x\leqslant n+k\leqslant n+k+1$, hence
$$
\log2-A_n\geqslant\sum_{k=1}^n\frac{1}{(n+k)(n+k+1)}\int_0^1(1-x)\mathrm dx.
$$
The integral is still $\frac12$ and the sum is
$$
\sum_{k=1}^n\left(\frac{1}{n+k}-\frac1{n+k+1}\right)=\frac1{n+1}-\frac1{2n+1}=\frac{n}{(n+1)(2n+1)},
$$
hence
$$
\log2-A_n\geqslant\frac{n}{2(n+1)(2n+1)}=\frac{u_n}{4n},
$$
with
$$
u_n=\frac1{(1+1/n)(1+1/(2n))}.
$$
Finally,
$\frac14u_n\leqslant n(\log2-A_n)\leqslant\frac14$ and $u_n\to1$ hence
$n(\log2-A_n)\to\tfrac14$.
