<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] 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.
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