What is the limit of the sequence $(a_n)_{n \ge 1}$, given that $1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} - a_n \ln n$ is bounded. Let $(a_n)_{n \ge 1}$ be a sequence of real numbers, such that the sequence:
$$1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} - a_n \ln n$$
(for $n \ge 1$) is bounded. What is the limit of the sequence $(a_n)_{n \ge 1}$?
I found this problem in a textbook and thought that it looks interesting. How should I approach something like this?
 A: Preliminaries
Defining the Harmonic Numbers
$$
H_n=\sum_{k=1}^n\frac1k\tag1
$$
and noticing that for $x\in[n,n+1]$, we have $\frac1{n+1}\le\frac1x\le\frac1n$, we get
$$
\overbrace{\ \ \frac1{n+1}\ \ \vphantom{\int^{n+1}}}^{H_{n+1}-H_n}\le\overbrace{\int_n^{n+1}\frac1x\,\mathrm{d}x}^{\log(n+1)-\log(n)}\le\overbrace{\quad\ \frac1n\quad\ \vphantom{\int^{n+1}}}^{H_n-H_{n-1}}\tag2
$$
The left-hand inequality shows that $H_{n+1}-\log(n+1)$ is decreasing.
The right-hand inequality shows that $H_n-\log(n+1)$ is increasing.
Together, these show that
$$
\gamma=\lim_{n\to\infty}(H_n-\log(n))\tag3
$$
exists and is between $H_0-\log(1)=0$ and $H_1-\log(1)=1$.

The Limit
We are given that
$$
H_n-a_n\log(n)=H_n-\log(n)-(a_n-1)\log(n)\tag4
$$
is bounded, say in $[L,M]$. Then, from what was said in the Preliminaries,
$$
\gamma\le H_n-\log(n)\le1\tag5
$$
and therefore,
$$
\frac{\gamma-M}{\log(n)}\le a_n-1\le\frac{1-L}{\log(n)}\tag6
$$
which, upon taking the limit, gives
$$
\lim_{n\to\infty}a_n=1\tag7
$$
A: Simple pair of diagrams shows this general fact:
if we have $f(x) > 0$ but $f'(x) < 0,$  then
$$ \int_a^{b+1} \; f(x) \;  dx \;  < \; \sum_{j=a}^b \;  f(j) \; < \; \int_{a-1}^b \;  f(x) \; dx   $$
Probably need to start with $a=2,$ with $b=n$ and $f(x) = \frac{1}{x}.$  Then add a $1$
$$ 1 + \int_2^{n+1} \; \frac{1}{x} \;  dx \;  < \; 1 + \sum_{j=2}^n \;   \frac{1}{j} \; < \;  1 + \int_{1}^n \;   \frac{1}{x} \; dx   $$ 
Often $\sum_{j=1}^n \frac{1}{j}$ is called $H_n$
A: You want
$\sum_{k=1}^n \dfrac1{k}
- a_n \ln n
=O(1)
$.
Since
$\sum_{k=1}^n \dfrac1{k}
=\ln(n)+O(1)
$,
you want
$\ln(n)+O(1)
- a_n \ln n
=O(1)
$
or
$(a_n-1)\ln(n)
=O(1)
$.
This requires
$a_n = 1+O(\dfrac1{\ln(n)})
$
so
$a_n \to 1$.
A: Your assumption is that the sequence $(u_n)_n$ defined by $$
u_n = H_n - a_n \ln n
$$
for $n\geq 1$ (where $H_n$ is the $n$-th Harmonic number) is bounded. I.e., there exists $C>0$ such that, for all $n\geq 1$,
$$
\lvert H_n - a_n \ln n \rvert \leq C
$$
Equivalently,
$$
\forall n\geq 2\,,\quad\left \lvert \frac{H_n}{\ln n} - a_n \right \rvert \leq \frac{C}{\ln n}
$$
and in particular,
$$
\lim_{n\to \infty }\left \lvert \frac{H_n}{\ln n} - a_n \right \rvert  = 0\,. \tag{1}
$$
But it is well-known that $H_n = \ln n + \gamma + o(1)$ (or, even then, $\displaystyle H_n \operatorname*{\sim}_{n\to\infty} \ln n$ suffices), i.e.,
$$
\lim_{n\to \infty }\frac{H_n}{\ln n} = 1\,. \tag{2}
$$
Combining (1) and (2) lets you conclude.
