Cauchy sequence to prove Integral does not exist

Question

I'm learning Cauchy sequence right now and didn't succeed.

prove that:

$$ \int_{i=1}^\infty \cfrac{x\,dx}{\sqrt{1+3x+x^2}} $$

Does not exist

Using only Cauchy sequence!

Thx!

Answer 1

Hint: Set $$x_n = \int_1^n\frac{xdx}{\sqrt{1+3x+x^2}}$$ and consider $|x_n-x_m|$.

Answer 2

Set $x_n:=\displaystyle{\int_{1}^{n}}\dfrac{x}{\sqrt{1+3x+x^2}}dx$

Note: For $x \ge 4:$

$x^2 > 1+3x$;

Let $\epsilon >0$ be given

Assume $x_n$ is Cauchy, i.e. there exists a $n_0$ s.t. for $m \ge n\ge n_0$

$|x_m-x_n| =\displaystyle{\int_{n}^{m}}\dfrac{x}{\sqrt{1+3x+x^2}}dx < \epsilon$.

Choose $n=\max (n_0,4)$, and $m=k+n$. For $k$ large enough we get

$\epsilon > |x_{k+n}-x_{n}|=$

$\displaystyle{\int_{n}^{k+n}}\dfrac{x}{\sqrt{1+3x+x^2}}dx>$

$\displaystyle{\int_{n}^{k+n}}\dfrac{x}{\sqrt{2x^2}}dx=$

$\displaystyle{\int_{n}^{k+n}}\dfrac{1}{√2}dx=\frac{1}{√2}k$, a contradiction.

Answer 3

As Dayton's hint. Let $\displaystyle a_n=\int_{1}^n \dfrac{x}{\sqrt{x^2+3x+1}}dx$ and definitely, assuming without loss of generality that $n\ge m$, $\displaystyle |a_n-a_m|=\left|\int_{1}^n\dfrac{x}{\sqrt{x^2+3x+1}}dx-\int_{1}^m\dfrac{x}{\sqrt{x^2+3x+1}}dx\right|=\int_{m}^n \dfrac{x}{\sqrt{x^2+3x+1}}dx$

We assume that $a_n$ converges as $n\rightarrow\infty$ and thus $\forall\epsilon>0$, $\exists N\in\mathbb{N_{\ge 1}}$ in which $|a_n-a_m|<\epsilon$ for any $n\ge m\ge N$. By choosing $n=m\sqrt{1+\dfrac{6}{m^2}}$ and $\epsilon = 1$ where $m\ge N\ge 1$. We obtain, as $\dfrac{x}{\sqrt{x^2+3x+1}}\ge \dfrac{1}{3}\Longleftrightarrow 3x(x-1)+(5x^2-1)\ge 0$ for all $x$ in the range $(m,n)$, $$1>\int_{m}^{n}\dfrac{x}{\sqrt{x^2+3x+1}}dx\ge\int_{m}^n\dfrac{dx}{3}=\dfrac{n^2-m^2}{6}=1$$ Which is such a contradiction $\Box$