Bounding the difference between $H_N$ and $\log N$ we consider the relation given: $$\int_n^{n+1}\frac{1}{x}dx < \frac{1}{n} < \int_n^{n+1}\frac{1}{x-1}dx$$
For $n>1$. We are to show $$\lim_{N\to\infty}\left[\left(\sum_{n=1}^N\frac{1}{n}\right) - \ln(N)\right]$$ is finite and to find the upper and lower bounds. My attempt is as follows. We note from the above inequality.: $$\sum_{n=1}^\infty\frac{1}{n}<\int_2^\infty\frac{1}{x-1}dx +1$$  Follows from here $$\lim_{N\to\infty}\left[\sum_{n=1}^\infty\frac{1}{n} - \ln(N)\right]<\lim_{N\to\infty}\left[\int_2^N\frac{1}{x-1}dx +1 - \ln(N)\right]$$ Which gives the upper bound as $1$. Similar logic for the lower bound gave the final result as $$0<\lim_{N\to\infty}\left[\left(\sum_{n=1}^N\frac{1}{n}\right) - \ln(N)\right]<1$$ Is this approach correct? Is it possible to have a tighter bound using the same approach? Many thanks
 A: Your argument correctly proves that
$$ S=\sum_{n=1}^{+\infty}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right) \tag{1}$$
converges to some constant in $(0,1)$. It is the Euler-Mascheroni constant $\gamma$:
$$ \gamma = \int_{0}^{+\infty}\left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right)e^{-x}\,dx = 0.5772156649\ldots\tag{2}$$
Notice that, from:
$$ H_N-\log(N+1) = \sum_{n=1}^{N}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right) = \gamma-\sum_{n>N}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right)\tag{3}$$
and from:
$$ \frac{1}{2n(n+1)}\leq\frac{1}{n}-\log\left(1+\frac{1}{n}\right) \leq \frac{1}{2n(n-1)},\tag{4} $$
that follows from the convexity of the function $g(t)=\frac{1}{t}$, we have:
$$ H_N = \log(N+1)+\gamma-\frac{1}{2N+\theta}, \qquad \theta\in[0,2],\tag{5}$$
that is way stronger than the initial bound. It is also possible to improve $(5)$ up to $\theta\in\left[\frac{16}{10},\frac{17}{10}\right]$, or up to:

$$ H_N = \log N + \gamma + \frac{1}{2N+\theta},\qquad \theta\in\left[\frac{1}{3},\frac{2}{5}\right].\tag{6}$$

