Problem with showing the inequality $\ln 2 -\frac{1}{2N}<\sum_{n=N+1}^{2N} \frac{1}{n}$ I have a problem with showing the following inequality
$$\ln 2 -\dfrac{1}{2N}<\sum\limits_{n=N+1}^{2N} \dfrac{1}{n}.$$
I had to show also that right side is smaller than $\ln 2$ but it is easily to conclude from inequality $\sum\limits_{k=2}^n \dfrac{1}{k}< \ln n$ (if I am not wrong). 
The bigger problem is with the inequality which I have written in the beginning. Unfortunately I have no ideas. 
I would appreciate any advice. Additionally I would like to ask if the limit
$$\lim_{N\to\infty} \sum\limits_{n=N+1}^{2N} \dfrac{1}{n}=0$$
is calculated properly, because result seems to be obvious.
 A: You can proceed in this way. Let
$$
\delta_N :=
\sum_{n=N+1}^{2N}\frac{1}{n} - \log(2) + \frac{1}{2N}.
$$
It is easy to prove that $(\delta_N)$ is a decreasing sequence.
Indeed, with some computation,
$$
\delta_{N+1} = \delta_N + \frac{1}{2N+1} - \frac{1}{2N} < \delta_N\,.
$$
Moreover, we have that
$$
\begin{align}
\delta_N & > \int_{N}^{2N-1} \frac{1}{x}\, dx -\log(2) + \frac{1}{2N} 
\\ & = \log\frac{2N-1}{N} -\log(2) + \frac{1}{2N}
= \log\left(1 - \frac{1}{2N}\right) + \frac{1}{2N}
\end{align}
$$
and the r.h.s. goes to $0$ as $N\to +\infty$.
By the previous step we conclude that $\lim_N \delta_N \geq 0$.
Hence, by monotonicity, $\delta_N > 0$ for every $N$.
A: Since $ x\mapsto\frac{1}{x} $ is decreasing on $ \mathbb{R}_{+}^{*}=\left(0,+\infty\right) $, we have for any $ k\in\mathbb{N}^{*} $ : $$ \frac{1}{k}=\int_{k}^{k+1}{\frac{\mathrm{d}x}{k}}\geq\int_{k}^{k+1}{\frac{\mathrm{d}x}{x}} $$
Thus, for $ N\in\mathbb{N} $, we have : $$ \sum_{k=N}^{2N-1}\frac{1}{k}\geq \sum_{k=N}^{2N-1}{\int_{k}^{k+1}{\frac{\mathrm{d}x}{x}}}=\int_{N}^{2N}{\frac{\mathrm{d}x}{x}}=\ln{2} $$
Hence : $$ \sum_{k=N+1}^{2N}{\frac{1}{k}}=-\frac{1}{N}+\sum_{k=N}^{2N-1}{\frac{1}{k}}+\frac{1}{2N}\geq\ln{2}-\frac{1}{2N} $$
A: $$\sum_{n=N+1}^{2N} \frac{1}{n}=H_{2 N}-H_N$$ Now, using, at least for large values of $N$, the expansion
$$H_p=\gamma+\log (p)+ \frac{1}{2 p}-\frac{1}{12   p^2}+O\left(\frac{1}{p^4}\right)$$
$$H_{2 N}-H_N=\log (2)-\frac{1}{4 N}+\frac{1}{16 N^2}+O\left(\frac{1}{N^4}\right)$$ making
$$\text{rhs - lhs }=\frac{1}{4 N}+\frac{1}{16 N^2}+O\left(\frac{1}{N^4}\right)$$
A: You may notice that
$$ H_{2N}-H_N=\sum_{k=N+1}^{2N}\frac{1}{k}=\sum_{k=1}^{2N}\frac{(-1)^{k+1}}{k}=\sum_{j=0}^{2N-1}\frac{(-1)^j}{j+1}=\int_{0}^{1}\sum_{j=0}^{2N-1}(-1)^j x^j\,dx $$
and this leads to
$$ H_{2N}-H_N = \int_{0}^{1}\frac{1-x^{2N}}{1+x}\,dx = \log(2)-\underbrace{\int_{0}^{1}x^{2N}\frac{dx}{1+x}}_{R(N)}. $$
Of course the remainder $R(N)$ is positive and bounded between $\int_{0}^{1}\frac{1}{2}x^{2N}\,dx=\frac{1}{4N+2}$ and $\int_{0}^{1}x^{2N}\,dx = \frac{1}{2N+1}$. A more accurate bound is given by
$$ R(N)=\frac{1}{4N+2}+\frac{1}{2}\int_{0}^{1}x^{2N}(1-x)\frac{dx}{1+x}$$
leading to
$$\frac{(4N+5)}{8(N+1)(2N+1)}\leq R(N)\leq \frac{(2N+3)}{4(N+1)(2N+1)}. $$
