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 :  
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.
 A: It is enough to show that $(\gamma_n)$ is decreasing and bounded below.
The function $f(x)=\dfrac{1}{x}$ is decreasing. On the interval $[k,k+1]$, $k\in\mathbb{N}$ we must have $\dfrac{1}{k+1}\leq f(x)\leq\dfrac{1}{k}$, hence
$$\dfrac{1}{k+1}\leq\int_k^{k+1}f(x)\,dx\leq\dfrac{1}{k}\hspace{1cm}(\ast)$$
(note that the length of the interval $[k,k+1]$ is $1$). Summing from $k=1$ to $n-1$ we get
$$\sum_{k=1}^{n-1}\dfrac{1}{k+1}\leq\int_1^n \dfrac{1}{x}\,dx\leq\sum_{k=1}^{n-1}\dfrac{1}{k}$$
that is
$$\sum_{k=1}^n\dfrac{1}{k}-1\leq\int_1^n\dfrac{1}{x}\,dx\leq\sum_{k=1}^n\dfrac{1}{k}-\dfrac{1}{n}$$
The RHS inequality gives us
$$\gamma_n=\sum_{k=1}^n\dfrac{1}{k}-\int_1^n\dfrac{1}{x}\,dx\geq\dfrac{1}{n}\geq 0$$
so that $(\gamma_n)$ is bounded below by $0$. To show that $(\gamma_n)$ is decreasing, consider
$$\gamma_{n}-\gamma_{n-1}=\dfrac{1}{n}-\int_{n-1}^n\dfrac{1}{x}dx\leq 0$$
by virtue of the LHS of $(\ast)$, so that $(\gamma_n)$ is decreasing.
A: Hint Consider the series $\sum_{n}(\gamma_{n}-\gamma_{n-1})$. It's convergent because
$$\gamma_{n}-\gamma_{n-1}=\frac1n+\ln\left(1-\frac1n\right)\sim-\frac12\frac1{n^2}$$
Moreover by telescoping we see that the sequence $(\gamma_n)_n$ is then convergent. 
A: 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.
A: The integral is $\ln n$
Consider the series
$$\sum \frac{1}{n}-\ln (1+\frac{1}{n})$$ the partial sums are equal to $\gamma_n$ so you want to show the series converges. This follows by comparison with $$\sum \frac{1}{n(n+1)}$$ using $$-\ln(1+\frac{1}{n})<-\frac{1}{n+1}$$
