<s>[INST] Increasing and bounded sequence proof Prove that the sequence $a_n= 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln(⁡n)$ is increasing and bounded above. Conclude that it’s convergent.
This what I got so far
Proof:
Part 1: Proving $a_n$ is increasing by induction.
Base: 
$a_1=1$
$a_2=1+\frac 12= \frac 32$
$a_1≤a_2$
So the base case is established.
Induction step: We assume that $a_{n-1}≤a_n$. We will show that $a_n≤a_{n+1}$.
Since 
$a_{n-1}≤a_n$
$$1+ \frac 12+ \frac 13+\cdots+ \frac{1}{(n-1)}-\ln(n-1) \leq 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln n$$
How should I continue?
 [/INST] @Patrick's answer shows that the sequence is decreasing.  However, we can show boundedness and covergence in a single step since the terms of the sequence are non-negative.  I only provide a skeleton solution, which will need fleshing out with a limit argument.
We can begin by rewriting the sequence as:
\begin{equation*} \sum_{k=1}^n \frac{1}{k} -\log{n} = \sum_{k=1}^n \frac{1}{k} -\int_{1}^n \frac{1}{x} d{x}
       = \overbrace{\sum_{k=1}^{n-1} \int_{k}^{k+1} \Bigl( \frac{1}{k}  -\frac{1}{x} \Bigr) d{x} +\frac{1}{n}}^\circledast.
\end{equation*}
The integral's limits provide the inequalities $k\le x \le {k+1} $, with which we obtain the following bounds on the integrand:
$$0 \le \frac{1}{k} - \frac{1}{x} \le  \frac{1}{k} -\frac{1}{k+1} = \frac{1}{k(k+1)} < \frac{1}{k^2}.$$  Then, after substituting our new integrand $\tfrac{1}{k^2}$, we obtain:
\begin{equation*} 0 \le \sum_{k=1}^n \frac{1}{k} -\log{n} =\overbrace{\sum_{k=1}^{n-1} \int_{k}^{k+1} \Bigl( \frac{1}{k}  -\frac{1}{x} \Bigr) d{x} +\frac{1}{n}}^\circledast \le \overbrace{\sum_{k=1}^{n-1}  \frac{1}{k^2}}^{\spadesuit} +\frac{1}{n}.
\end{equation*}
By comparison with the convergent series $\spadesuit$, we find that the original sequence converges.
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