<s>[INST] 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?
 [/INST] 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, $$ n(\log 2 - A_n) \to \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.$$
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