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???

An elementary way to see the inequality is to recall the formal definition of $e^x=1+x+\frac{x^2}{2!}+\ldots$, so using $\log(n+1)-\log(n)=\log(1+1/n)$ and $1+1/n\leq e^{1/n}$ gives the result upon taking logarithms. The lower bound follows similarly. 
Using the inequality your professor gave, look at $a_{n+1}-a_n$ and you'll see the sum term becomes very simple. 
