# Use comparison test to see if series converges

• kwal0203
In summary: Well do something with that fact!Oh! lnx < x iff e^(lnx) < e^x iff x < e^x.So (lnx)^9 < e^x, and we know e^x diverges so it must be that (lnx)^9 also diverges.In summary, we can use the fact that e^x eventually becomes greater than x^n for all n, and the inverse relationship between ln and exp, to show that (lnx)^9 also diverges.
kwal0203

## Homework Statement

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

## 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!

kwal0203 said:
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!

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##.

PeroK said:
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##.

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:

$$(lnx)^9 < x$$

?

kwal0203 said:
Well the condition on x to make that true is x > 1 right?

How do you prove that?

PeroK said:
How do you prove that?

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.

kwal0203 said:
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.

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?

PeroK said:
Okay. But can you extend that idea to ##(\ln x)^2##?

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.

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

Another way to prove lnx < x?

kwal0203 said:
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?

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

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

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?

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.

PeroK said:
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.

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)

kwal0203 said:
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)

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.

PeroK said:
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.

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.

kwal0203 said:
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.

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

PeroK said:
You know that ##\ln## is the inverse of ##\exp##?

Yes

kwal0203 said:
Yes

Well do something with that fact!

## 1. How does the comparison test determine if a series converges?

The comparison test compares the given series to another series whose convergence is already known. If the known series converges and the given series is less than or equal to it, then the given series also converges. If the known series diverges and the given series is greater than or equal to it, then the given series also diverges. If the comparison is inconclusive, then the given series may or may not converge.

## 2. Can the comparison test be used for all series?

No, the comparison test can only be used for series with non-negative terms. If the series has negative terms, then the absolute value of the terms must be used for comparison.

## 3. What is the difference between the comparison test and the limit comparison test?

The comparison test compares the given series to a known series, while the limit comparison test compares the ratios of the terms of the given series and the known series. Both tests have the same result, but the limit comparison test is more useful when the two series have different orders of magnitude.

## 4. Is the comparison test a sufficient condition for convergence?

Yes, the comparison test is a sufficient condition for convergence. If the given series satisfies the conditions of the comparison test, then it must converge. However, if the given series does not satisfy the conditions, it may still converge or diverge.

## 5. Can the comparison test be used to determine the exact value of the sum of a series?

No, the comparison test only determines if a series converges or diverges. It does not provide information about the exact value of the sum. Other tests, such as the ratio test or the integral test, can be used to approximate the sum of a convergent series.

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