Prove that $\{n\}$ is a Cauchy sequence that doesn't converge. 
Consider the distance function given by $d:\mathbb{R}\times\mathbb{R}\to\mathbb{R},\;d(x,y)=|\int_x^yf(t)dt|$ where $f$ is a continuous and positive function such that $\int_{-\infty}^{+\infty}f$ converges. Show that $\{n\}$ is  Cauchy sequence that doesn't converge.

My idea to solve this: Suppose $\{n\}$ is not a Cauchy sequence, then there is no $N$ such that $d(n,n+1)<\epsilon$ if $n>N$ for any given $\epsilon>0$. I want to show how this assumption contradicts the fact $\int_{-\infty}^{+\infty}f<\infty$.
Since we suppose $\{n\}$ is not Cauchy, this means that there are infinitely many intervals of the form $(n_i,n_i+1)$ such that $\int_{n_i}^{n_i+1}f\geq\epsilon$ (and there must be infinite many intervals, otherwise we would have a last one and then $\{n\}$ would be Cauchy).
For any given $K$ pick $H$ integer such that $K\leq\epsilon H$, then take $H$ intervals $(n_i,n_i+1)$ that makes $\int_{n_i}^{n_i+1}f\geq\epsilon$. Now we have $H$ intervales, let those intervals be $(n_1,n_1+1),\dots,(n_H,n_{H}+1)$.
Using the fact that $f$ is always positive we have $|\int_{n_1}^{n_{H+1}} f(t)dt|<|\sum_{i=1}^{H}\int_{i}^{i+1}f(t)dt|\leq \sum_{i=1}^{H}|\int_{i}^{i+1}f(t)dt|\leq \sum_{i=1}^{H} \epsilon =\epsilon H\geq K$.
Then if $\{n\}$ is not Cauchy is possible to find limits of integration such that the integral is bigger than $K$ for any given $K$, this contradicts $\int_{-\infty}^{+\infty} f\leq \infty$.
Is this part of the proof correct?.
Now I have to prove that $\{n\}$ doesn't converge, but I'm not sure how to do this part.My idea here was that if I suppose the sequence converge to some $x$ then is possible find $n$ such that $d(n,x)<\epsilon$ but also $d(x,x+1)<\epsilon$ and $d(x+1,x+2)<\epsilon$ then the sequence doesn't converge to $x$.

$\langle 1, 2, 3, \ldots \rangle$ is Cauchy because the $\lim_{n\rightarrow\infty} \int_n^\infty f(x) dx =0$ and for any $N>n$ we have $\int_n^\infty f(x) dx \ge \int_n^{N}f(x)dx$. 
$\langle 1, 2, 3, \ldots \rangle$ is not convergent because given any $L$ we may pick $n$ such that $n>L$. Then $\int_L^n f(x)dx=\epsilon>0$, since $f$ is strictly positive. But $\int_L^N f(x)dx \ge \int_L^n f(x)dx$ for all $N>n$. So $d(L,N)>\epsilon$ for all $N>n$. 
