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?
 A: This sequence is NOT increasing and is in fact DECREASING.
Proof:
$$a_n = 1+\frac{1}{2} + \cdots + \frac{1}{n} - \ln(n)$$
$$a_{n+1}=1+\frac{1}{2} + \cdots + \frac{1}{n}+\frac{1}{n+1} - \ln(n+1)$$
so 
$$a_{n+1}=a_n + (\ln(n) +\frac{1}{n+1} - \ln(n+1)) = a_n + \frac{1}{n+1} - (\ln(n+1) - \ln(n))$$
$$\ln(n+1)-\ln(n) = \int_n^{n+1}\frac{1}{t}dt > \frac{1}{n+1}$$
To see the last line just draw a graph of $\frac{1}{t}$ between $n$ and $n+1$ and draw a box of width $1$ and height $\frac{1}{n+1}$ (i.e. the box touches $\frac{1}{t}$ at $t=\frac{1}{n+1}$).
NOTE: Your base case was not done properly since you forgot to subtract $\ln(1)$ and $\ln(2)$ (granted $\ln(1) = 0$) if you subtracted $\ln(2) \approx 0.693147 > 0.5$ you would have seen the base case not to hold.
A: I think you should have been asked to show that
$$
1+\frac12+\frac13+\cdots+\frac1n-\ln(n+1) \tag 1
$$
increases with $n$.  The previous term is
$$
1+\frac12+\frac13+\cdots+\frac{1}{n-1} - \ln n. \tag 2
$$
Then the problem is to show that when you subtract $(2)$ from $(1)$, the difference is nonnegative.  You get
$$
\frac1n - \ln(n+1) + \ln n = \int_n^{n+1} \frac1n - \frac1x \, dx. \tag 3
$$
It is easy to show that that is nonnegative.
To show that $(1)$ is bounded above, first notice that $(1)$ is equal to the sum of expressions like the one in $(3)$:
$$
\sum_{k=1}^n \int_k^{k+1} \frac1k - \frac1x\, dx. \tag 4
$$
Then
$$
[\text{expression in $(4)$}] \ge \sum_{k=1}^n \int_k^{k+1} \frac1k - \frac{1}{k+1}\, dx = \sum_{k=1}^n \frac1k - \frac{1}{k+1},
$$
and this is a telescoping sum, and all its terms are non-negative, and it adds up to $1$.
A: @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.
