Show that monotone sequence is reducing and bounded by zero? Let $a_n=\displaystyle\sum\limits_{k=1}^n {1\over{k}} - \log n$ for $n\ge1$. Euler's Constant is defined as $y=\lim_{n\to\infty} a_n$. Show that $(a_n)^\infty_{n=1}$ is decreasing and bounded by zero, and so this limit exists

My thought:
When I was trying the first few terms for $a_1, a_2, a_3$, I get:
$$a_1 = 1-0$$
$$a_2={1\over1}+{1\over2}-\log2$$
$$a_3={1\over1}+{1\over2}+{1\over3}-\log3$$
It is NOT decreasing!!! what did I do wrong??
Professor also gave us a hint: Prove that $\dfrac{1}{n+1}\le \log(n+1)-\log n\le \dfrac1n$
we need to use squeeze theorem???

Certainly it's decreasing.  I get $1-0=1$, and $1 + \frac12 -\log2=0.8068528\ldots$, and $1+\frac12+\frac13-\log3=0.73472\ldots$.
$$
\frac{1}{n+1} = \int_n^{n+1}\frac{dx}{n+1} \le \int_n^{n+1}\frac{dx}{x} = \log(n+1)-\log n. 
$$
$$
\frac1n = \int_n^{n+1} \frac{dx}{n} \ge \int_n^{n+1} \frac{dx}{x} = \log(n+1)-\log n.
$$
So
$$
\Big(1+\frac12+\frac13+\cdots+\frac1n-\log n\Big) +\Big(\log n\Big) + \Big(\frac{1}{n+1}-\log(n+1)\Big)
$$
$$
\le1+\frac12+\frac13+\cdots+\frac1n-\log n.
$$
