TITLE: Prove that for each $n \in \mathbb N$, $\gamma_n=1+\frac{1}{2}+\dots+\frac{1}{n}-\int_1^n \frac{1}{x}dx$ is convergent
QUESTION [5 upvotes]: Question :  
For each $n \in \mathbb N$, Define :
$\gamma_n=1+\frac{1}{2}+\dots+\frac{1}{n}-\int_1^n \frac{1}{x}dx$  
Prove that $\{\gamma_n\}$ is convergent.    
Note 1 :  I know that $\sum_1^\infty \frac{1}{k}$ diverges. I can guess that this sequence is convergent to $0$ because of the similar terms inside the integration and outside of it.  The problem is that the integration is continuous. I mean that $x$ is not just integers.  What can we do with the values left?! ( For example, $\frac{1}{2.5}$ ) Also, Notice that this problem should be solved with some elementary tools.  ( It's from a part of my book which is even before any discussion about the Fundamental theorem of calculus. ) 
Note 2 : I'm not good at integration and i'm learning it. Any good details can help me.  
Thanks in advance.

REPLY [2 votes]: Note that
$$
\begin{align}
\gamma_{k-1}-\gamma_k
&=\int_{k-1}^k\frac1x\,\mathrm{d}x-\frac1k\\
&=\int_{k-1}^k\left(\frac1x-\frac1k\right)\mathrm{d}x\\[9pt]
&\gt0
\end{align}
$$
Thus, $\gamma_k$ is a decreasing sequence. Furthermore,
$$
\begin{align}
\gamma_{k-1}-\gamma_k
&=\int_{k-1}^k\frac1x\,\mathrm{d}x-\frac1k\\
&=\int_{k-1}^k\left(\frac1x-\frac1k\right)\mathrm{d}x\\
&=\int_{k-1}^k\frac{k-x}{kx}\,\mathrm{d}x\\
&\le\frac1{k(k-1)}
\end{align}
$$
Therefore,
$$
\begin{align}
\gamma_1-\gamma_n
&=\sum_{k=2}^n\left(\gamma_{k-1}-\gamma_k\right)\\
&\le\sum_{k=2}^n\left(\frac1{k-1}-\frac1k\right)\\
&=1-\frac1n
\end{align}
$$
That is, $\gamma_n$ is bounded below by $\gamma_1-1=0$.
So, $\gamma_n$ is decreasing and bounded below by $0$. Therefore, $\lim\limits_{n\to\infty}\gamma_n\ge0$ exists.