Need some help here.
I need prove that the following limit exist and is less than $1$
$$\lim_{x\to\infty}\left(\sum_{n\le x}^{}\frac{1}{n}-\ln x\right)$$
I feel a little lost here, this is my first exercise of this type, and I don't know what to use.
Any help is welcome!
This is a more general claim, but the proof of what you have is the same, just that $f$ is in incognito now.
Suppose that $f:\Bbb R\to\Bbb R$ is a positive, monotone decreasing function such that $\lim\limits_{x\to\infty}f(x)=0$. Define $$s_n=\sum_{k=1}^n f(k)$$ $$t_n=\int_1^n f(x) dx$$ $$d_n=s_n-t_n$$
Then $\displaystyle\lim\limits_{n\to\infty}d_n=D$ exists, and $0\leq D\leq f(1)$ and $\displaystyle \sum_{k=1}^\infty f(k)$ exists if and only if $\displaystyle \int_1^\infty f(x)dx$ does.
NOTE The last claim is something you will find useful, but it does not address your question.
PROOF We have that $$\tag{ 1}t_{n+1}=\int_1^{n+1}f(x)dx=\sum_{k=1}^n \int_k^{k+1}f(x)dx\color{green}{\leq} \sum_{k=1}^n \int_k^{k+1}f(k)dx=\sum_{k=1}^n f(k)=s_n$$
which means $$0\leq f(n+1)=s_{n+1}-s_n\color{red}{\leq} s_{n+1}-t_{n+1}=d_{n+1}$$ where we used $\color{red}{(1)}$, so that $$0\leq d_{n+1}$$
Next, $$d_n-d_{n+1}=\int_n^{n+1} f(x) dx-f(n+1)\color{green}{\geq} \int_n^{n+1} f(n+1) dx-f(n+1)=0$$ so $d_n$ is decreasing. By the montone convergence theorem, $D$ exists. Since $d_1=f(1)$, it is proven that $$0\leq D\leq f(1)$$
Pay attention to where the fact that $ {f\geq 0}$ and that $\color{green}{f \text{ is decreasing}}$ has been used. A drawing should prove very, very useful!
Now use the above with $f(x)=\dfrac 1 x $.
ADD This, for example, proves with little machinery that $$\lim_{n\to\infty}\left(2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt k}\right)$$
exists and is $1<\ell<2$. The limit is $-\zeta(1/2)$ (credits to Rob), which shows its evaluation is anything but trivial.
Another approach: define
$$f(x):=\log(1-x)+x\;,\;x<1\implies f'(x)=-\frac1{1-x}+1=0\iff x=0\;,\;\;\text{and since}$$
$$f''(x)=-\frac1{(1-x)^2}<0\;\;\;\forall\,x<1\implies \,(0,0)\;\;\text{is a maximal point of}\;\; f(x)\implies\,$$
$$\forall\,x<1\;,\;\;f(x)<f(0)=0\;\;\;\;\;\color{red}{(*)}$$
Now define
$$H_n:=\sum_{k=1}^n\frac1n\;,\;\;\text{and observe that}\;\;H_{n+1}-H_n=\frac1{n+1}\;,\;\;\text{so if we define}$$
$$a_n:=H_n-\log n\implies a_{n+1}-a_n=\frac1{n+1}-\log(n+1)+\log n=\log\left(1-\frac1{n+1}\right)+\frac1{n+1}$$
and from $\;\color{red}{(*)}\;$ above we get that $\,a_{n+1}<a_n\;$ , so the sequence $\,\{a_n\}\,$ is monotone descending.
OTOH, taking the upper Riemann sum for the function $\,\displaystyle{f(x):=\frac1x}\,$ of the partition $\,\{x_1=1\,,\,x_2=2\,,\ldots\,,x_n=n\}\,$ of the interval $\,[1,n]\,$ (i.e., choosing the left extreme points of every subinterval as the function $\;f\;$ is monotone descending), we get
$$\log n=\int\limits_1^n\frac{dx}x\le\sum_{k=1}^{n-1}\frac1{x_k}(x_{k+1}-x_k)=\sum_{k=1}^{n-1}\frac1k=1+\frac12+\ldots+\frac1{n-1}=H_n-\frac1n$$
so with the first part above we get
$$a_n=H_n-\log n\ge\frac1n$$
and we thus get the sequence $\,\{a_n\}\,$ is monotone descending and bounded from below and thus its limit exists, which we call "The Euler-Mascheroni Constant" and denote it by $\,\gamma\,$ ...
Note: The part after "OTOH" can be made pretty easy and compelling with a diagram showing the function $\,f\,$ and the "upper" rectangles for the corresponding integral.
