# Test for Convergence

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Suppose I want to test for the

How would I go about it, does anyone have any ideas?

I was thinking that:

and that

but this doesn't help because the integral for the LHS converges and the integral for the RHS diverges and so the direct comparison test definitely doesn't help here.

I thought since for x near 0, then can I say that and so for x near 0 and so and as that converges, then by the DCT, so does our integral absolutely?

**absolute**convergence of .How would I go about it, does anyone have any ideas?

I was thinking that:

and that

but this doesn't help because the integral for the LHS converges and the integral for the RHS diverges and so the direct comparison test definitely doesn't help here.

I thought since for x near 0, then can I say that and so for x near 0 and so and as that converges, then by the DCT, so does our integral absolutely?

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#2

(Original post by

Suppose I want to test for the

How would I go about it, does anyone have any ideas?

I was thinking that:

and that

but this doesn't help because the integral for the LHS converges and the integral for the RHS diverges and so the direct comparison test definitely doesn't help here.

I thought since for x near 0, then can I say that and so for x near 0 and so and as that converges, then by the DCT, so does our integral absolutely?

**Takeover Season**)Suppose I want to test for the

**absolute**convergence of .How would I go about it, does anyone have any ideas?

I was thinking that:

and that

but this doesn't help because the integral for the LHS converges and the integral for the RHS diverges and so the direct comparison test definitely doesn't help here.

I thought since for x near 0, then can I say that and so for x near 0 and so and as that converges, then by the DCT, so does our integral absolutely?

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(Original post by

Over the interval (0,1) we have |sin(x)| < x so the second approach is correct.

**RDKGames**)Over the interval (0,1) we have |sin(x)| < x so the second approach is correct.

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#4

(Original post by

Thank you, why did you talk about an open interval rather than (0,1], just wondering.

**Takeover Season**)Thank you, why did you talk about an open interval rather than (0,1], just wondering.

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(Original post by

Does it matter?

**RDKGames**)Does it matter?

Also, just wondering if my upper limit of the integral was x = 2 pi, then would this still hold?

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#6

(Original post by

I don't know, that's why i asked.

Also, just wondering if my upper limit of the integral was x = 2 pi, then would this still hold?

**Takeover Season**)I don't know, that's why i asked.

Also, just wondering if my upper limit of the integral was x = 2 pi, then would this still hold?

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(Original post by

If I asked you to integrate something over the region (0,1) and then ask you to integrate over [0,1] instead, how is your working out going to be different?

**RDKGames**)If I asked you to integrate something over the region (0,1) and then ask you to integrate over [0,1] instead, how is your working out going to be different?

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#8

(Original post by

I don't think it will, I'll do it the same way.

**Takeover Season**)I don't think it will, I'll do it the same way.

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(Original post by

So it doesnt matter whether I talk about the open interval or the closed one, or the half-open one.

**RDKGames**)So it doesnt matter whether I talk about the open interval or the closed one, or the half-open one.

Would I have to split the integral from x = 0 to x = 1 and use the above method and then say from x = 1 to x = 2pi, the integrand is continuous and well defined so the integral is 'proper' and converges.

So as both integrals converge, the integral from x = 0 to x = 2pi does too?

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#10

(Original post by

Thank you, would my above approach still work if Also, just wondering if my upper limit of the integral was x = 2 pi, then would this still hold?

Would I have to split the integral from x = 0 to x = 1 and use the above method and then say from x = 1 to x = 2pi, the integrand is continuous and well defined so the integral is 'proper' and converges.

So as both integrals converge, the integral from x = 0 to x = 2pi does too?

**Takeover Season**)Thank you, would my above approach still work if Also, just wondering if my upper limit of the integral was x = 2 pi, then would this still hold?

Would I have to split the integral from x = 0 to x = 1 and use the above method and then say from x = 1 to x = 2pi, the integrand is continuous and well defined so the integral is 'proper' and converges.

So as both integrals converge, the integral from x = 0 to x = 2pi does too?

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(Original post by

Yep

**RDKGames**)Yep

Just wondering, when i take Taylor series of sin x about x = 0, how do I know that sin x <= x for x in (0,1).

Like how would I know for what values around x = 0 will the inequality hold?

It just seems to hold everytime I have an integral like here from 0 to 1, but from 0 to 5, it'd be different etc.

Last edited by Takeover Season; 6 months ago

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#12

(Original post by

thanks, so I'd definitely be wrong in saying that sin x <= x over x in (0,2pi) right?

**Takeover Season**)thanks, so I'd definitely be wrong in saying that sin x <= x over x in (0,2pi) right?

< is correct but so is the less strict <=

Just wondering, when i take Taylor series of sin x about x = 0, how do I know that sin x <= x for x in (0,1).

Just sketch the graphs of sin x and x.

x > sin x for +ve x

x < sin x for -ve x

Last edited by RDKGames; 6 months ago

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(Original post by

No need to consider Taylor. Just sketch the graphs.

x > sin x for +ve x

x < sin x for -ve x

**RDKGames**)No need to consider Taylor. Just sketch the graphs.

x > sin x for +ve x

x < sin x for -ve x

i think what i was meant to say is, when i try to find the behaviour of sin x near x = 0, the problem point of the integral.

i use the taylor series to find this of sin x about x = 0 to find this.

then i see that sin x <= x for this.

i guess after knowing this, I'd just draw sin x and x because they're simple and see for what x does the inequality hold.

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#14

(Original post by

oh right, thank you.

i think what i was meant to say is, when i try to find the behaviour of sin x near x = 0, the problem point of the integral.

i use the taylor series to find this of sin x about x = 0 to find this.

then i see that sin x <= x for this.

i guess after knowing this, I'd just draw sin x and x because they're simple and see for what x does the inequality hold.

**Takeover Season**)oh right, thank you.

i think what i was meant to say is, when i try to find the behaviour of sin x near x = 0, the problem point of the integral.

i use the taylor series to find this of sin x about x = 0 to find this.

then i see that sin x <= x for this.

i guess after knowing this, I'd just draw sin x and x because they're simple and see for what x does the inequality hold.

You may even search the web for its proof since its not too difficult.

As I said, Taylor expansion was just to get you thinking of the behaviour of the integrand near the bad point, and is often more useful in more complicated integrands.

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(Original post by

I suggest you recall this inequality regarding x and sin x since its can slap it on many integrals to show their convergence.

You may even search the web for its proof since its not too difficult.

As I said, Taylor expansion was just to get you thinking of the behaviour of the integrand near the bad point, and is often more useful in more complicated integrands.

**RDKGames**)I suggest you recall this inequality regarding x and sin x since its can slap it on many integrals to show their convergence.

You may even search the web for its proof since its not too difficult.

As I said, Taylor expansion was just to get you thinking of the behaviour of the integrand near the bad point, and is often more useful in more complicated integrands.

Yes, that makes sense!

I was just wondering that does the Taylor Expansion usefulness diminish when testing for absolute convergence? If I want to find the Taylor Expansion of |f(x)|, do I just find the Taylor Expansion of f(x) and then take the absolute value of both sides?

I was just wondering because if I can't do that, then it would be difficult, especially if f(x) <= h(x) for some interval from the Taylor Expansion, but then the findings doesn't hold when we take |f(x)|.

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