Prove that $\{n\}$ is a Cauchy sequence that doesn't converge.

Question

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 $x$.

Answer 1

$\langle 1, 2, 3, \ldots \rangle$ 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 $N>n$.

Answer 2

$a_n=n$ is Cauchy, since $$\sum_{n\geqslant 0} d(x_n,x_{n+1})=\int_0^\infty f<+\infty$$

Answer 3

Here's a proof that the sequence doesn't converge. Fix $x \in \Bbb R$, and consider the sequence $d_n=d(x,n)$. Pick the smallest $n$ greater than $x$; call it $n_x$. Then $f$ has a positive minimum on the interval $[x,n_x]$ by compactness (aka, by the extreme value theorem.) Call this minimum $m$. Then $$\int_x^{n_x} f(t)dt \geq m(x-n_x)$$ By the triangle inequality, if $n>n_x$, then $d_n \geq d_{n_x}$, which is a positive real number. Thus $\limsup d_n \geq d(x,n_x) >0$, so the sequence does not converge to $x$. As $x$ was arbitrary the sequence does not converge.

Answer 4

Lets imagine ${n}$ converges to $x$, then lets suppose $x<n_0$ for some natural number $n_0$, then lets take $$k=\int_x^{n_0}f(t)dt$$

we have that $$\int_x^{n_0}f(t)dt \leq \int_x^nf(t)dt$$ for every $n>n_0$ (since $f$ is positive). Now take $\epsilon< k$, since ${n}$ converges to $x$ there exists an $N$ such that for every $n>N$ we have $$\int_x^{n}f(t)dt < \epsilon $$ taking $n>\max\{n_0,N\}$ we get that $$k=\int_x^{n_0}f(t)dt \leq \int_x^nf(t)dt <\epsilon$$ which is a contradiction, this means $x\geq n_0$. We can do this for any natural number, which means $x>n$ for all $n\in \mathbb{N}$, once again a contradiction.

The reasoning behind this demonstration comes from looking the sequence in the real line and realizing that the only limit "posible" would be at infinity.

Answer 5

Since $$\biggr|\int_{0}^{\infty}f(x)dx \biggr|<\infty,$$ given $\varepsilon >0$ exists $N$ natural such that $$\biggr|\int_{N}^{\infty} f(x)dx\biggr|<\varepsilon.$$ Since $f$ is positive, it is obvious that $$\biggr| \int_{N+1}^{\infty}f(x)dx\biggr|< \varepsilon.$$ Using the triangle inequality, $$\biggr|\int_{N}^{N+1} f(x)dx \biggr| <2 \varepsilon.$$ Then, the sequence is indeed a Cauchy sequence. Now, let us prove that it does not converge. Suppose that there is a real number $a$ such that $\lim d(n, a)=0$. Then, we would have $$\lim \biggr| \int_{n}^{a} f(x)dx \biggr|=0.$$ But this statement says that if $m$ is any natural number, then $$d(m, a) \leq \lim d(n, a)=0.$$ Since $f$ is continuous and positive, we would have $a=m$ for any natural number $m$. This contradiction shows that we could not have a limit for that sequence. In another words, $\mathbb{R}$ is not complete with this metric.