Determining whether an integral converges based on comparison

[tmt1]

I have this integral

$$\int_{1}^{\infty} \frac{1}{(x + x^2)^\frac{1}{3}} \,dx$$

I think this is smaller than

$$\int_{1}^{\infty} \frac{1}{(x^2)^\frac{1}{3}} \,dx$$

The latter converges since $p = \frac{2}{3}$ which is greater than 1. Therefore, since the first integral is smaller than the second integral it must also converge. However, the answer is that it diverges. What is the error in my logic?

 

[MarkFL]

Well for starters, is it really true that \(\displaystyle \frac{2}{3}>1\)?

We know that:

\(\displaystyle \int_1^{\infty}\frac{1}{x}\,dx\) diverges. So, what we can do is try to show that:

\(\displaystyle \left(x+x^2\right)^{\frac{1}{3}}\le x\) where \(\displaystyle 1\le a\le x\)

 

[tmt1]

Well for starters, is it really true that \(\displaystyle \frac{2}{3}>1\)?

We know that:

\(\displaystyle \int_1^{\infty}\frac{1}{x}\,dx\) diverges. So, what we can do is try to show that:

\(\displaystyle \left(x+x^2\right)^{\frac{1}{3}}\le x\) where \(\displaystyle 1\le a\le x\)

If I raise each side to the power of 3, I get:

$$x + x^2 \le x^3$$

Is that enough proof that \(\displaystyle \left(x+x^2\right)^{\frac{1}{3}}\le x\)?

 

[MarkFL]

If I raise each side to the power of 3, I get:

$$x + x^2 \le x^3$$

Is that enough proof that \(\displaystyle \left(x+x^2\right)^{\frac{1}{3}}\le x\)?

Since $1<x$, we can divide through by $x$ and then arrange as follows:

\(\displaystyle x^2-x-1\ge0\)

Where is this true?

 

[tmt1]

Since $1<x$, we can divide through by $x$ and then arrange as follows:

\(\displaystyle x^2-x-1\ge0\)

Where is this true?

So

\(\displaystyle x^2-x\ge 1\)

and

\(\displaystyle x(x - 1)\ge 1\)

so it looks like this is true whenever x is greater than 1.

 

[MarkFL]

What about \(\displaystyle x=\frac{3}{2}\)?

What are the roots of $x^2-x-1=0$?

 

[tmt1]

Since $1<x$, we can divide through by $x$ and then arrange as follows:

\(\displaystyle x^2-x-1\ge0\)

Where is this true?

I'm having trouble reasoning about this.

We need to make this assertion $(x + x^2)^{\frac{1}{3}} \le x$ where $1 \le a \le x$. (What is $a$ here, by the way?).

Since $x > 1$, we can do:

$ \frac{(x + x^2)^{\frac{1}{3}}}{x} \le \frac{x}{x}$

Or $ \frac{(x + x^2)^{\frac{1}{3}}}{x} \le 1$

But how can we evaluate $\frac{(x + x^2)^{\frac{1}{3}}}{x}$?