# Use comparison test to see if series converges

1. Jan 21, 2017

### kwal0203

1. The problem statement, all variables and given/known data

$$\sum_{x=2}^{\infty } \frac{1}{(lnx)^9}$$

2. Relevant equations

3. The attempt at a solution
$$x \geqslant 2$$

$$0 \leqslant lnx < x$$
$$0 < \frac{1}{x} < \frac{1}{lnx}$$

From this we know that 1 / lnx diverges and I wanted to use this fact to show that 1 / [(lnx) ^ 9] diverges but at k >= 2 we have this:

$$0 < \frac{1}{(lnx)^9} < \frac{1}{lnx}$$

So it doesn't really work.

Is there a relation between 1/x and 1/[(lnx)^9] that I can use to solve the problem?

Any help would be appreciated, thanks!

2. Jan 21, 2017

### PeroK

Hint: work out how to show the condition on $x$ such that $\ln x < x$ and see whether you can extend this for $(\ln x)^9$.

3. Jan 21, 2017

### kwal0203

Well the condition on x to make that true is x > 1 right?

To extend it, should I try and find when this is true:

$$(lnx)^9 < x$$

?

4. Jan 21, 2017

### PeroK

How do you prove that?

5. Jan 21, 2017

### kwal0203

Well, lnx <= 0 whenever 0 < x <= 1 so that bit is obvious I guess.

Also, whenever x > 1, lnx < x because it's a monotonically increasing function with first derivative:

1 - 1/x > 0.

For that to be true x > 1.

6. Jan 21, 2017

### PeroK

Okay. But can you extend that idea to $(\ln x)^2$?

Or, is there another way to do it that might be easier to extend?

7. Jan 21, 2017

### kwal0203

I don't think you can because if you take lnx where x < 1, it's a negative result. So (lnx)^2 would be positive.

Another way to prove lnx < x?

8. Jan 21, 2017

### PeroK

Yes, just think of what you could do to that equation.

9. Jan 21, 2017

### kwal0203

Oh right,

$$lnx < \sqrt{x}$$
$$\sqrt{x} - lnx > 0$$
$$\frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{x}-lnx)=\frac{1}{2\sqrt{x}}-\frac{1}{x} > 0$$
when x >= 2.

so if that's true then so is this:
$$(lnx)^2 < x$$

Is that right?

10. Jan 21, 2017

### PeroK

Let me help. It seemed obvious to me:

$\ln x < x$ iff $x < e^x$

Perhaps that's easier to extend to powers of the logarithm.

11. Jan 21, 2017

### kwal0203

Ok, so we can say that $\ln x < e^x$ so $(\ln x)^9 < e^9x$ and $1/e^9x < 1/(lnx)^9$

Is that the idea? (those x's should be in the exponent as well)

12. Jan 21, 2017

### PeroK

That maths is all awry!

Generally, you can use the following result without proof:

For all $n$ eventually $e^x$ is greater than $x^n$. In fact the limit $\frac{e^x}{x^n}$ is $+\infty$ as $x \rightarrow +\infty$

That result should help in this case.

13. Jan 21, 2017

### kwal0203

Ok, yeah I got no idea.

I'm pretty sure 1/x < 1/(lnx)^9, I just can't prove it like I can for 1/x < 1/lnx.

And since 1/x diverges so does 1/(lnx)^9.

I'm not sure how the facts that lnx < x iff x < e^x (is this ever false?) and limit of e^x / x^n -> infinity helps.

14. Jan 21, 2017

### PeroK

You know that $\ln$ is the inverse of $\exp$?

15. Jan 21, 2017

### kwal0203

Yes

16. Jan 21, 2017

### PeroK

Well do something with that fact!