# Does this convergence? Justify your answer..

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An=(n^2+2^n)/((n^2)+2)

2 mark question from University past paper which is bugging me slightly.

I say that An is a sequence of non-negative integers and is not bounded, therefore, An diverges.

Is that enough of a justification or am I missing something here?

2 mark question from University past paper which is bugging me slightly.

I say that An is a sequence of non-negative integers and is not bounded, therefore, An diverges.

Is that enough of a justification or am I missing something here?

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#3

(Original post by

An=(n^2+2^n)/((n^2)+2)

2 mark question from University past paper which is bugging me slightly.

I say that An is a sequence of non-negative integers and is not bounded, therefore, An diverges.

Is that enough of a justification or am I missing something here?

**MSI_10**)An=(n^2+2^n)/((n^2)+2)

2 mark question from University past paper which is bugging me slightly.

I say that An is a sequence of non-negative integers and is not bounded, therefore, An diverges.

Is that enough of a justification or am I missing something here?

You basically need to be comparing the growth of the numerator with that of the denominator, which should be fairly easy to do since the n^2 terms are the same so you just need to say something about 2^n compared to 2.

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#4

(Original post by

How do you know the terms are integers - I don't think they are!

You basically need to be comparing the growth of the numerator with that of the denominator, which should be fairly easy to do since the n^2 terms are the same so you just need to say something about 2^n compared to 2.

**davros**)How do you know the terms are integers - I don't think they are!

You basically need to be comparing the growth of the numerator with that of the denominator, which should be fairly easy to do since the n^2 terms are the same so you just need to say something about 2^n compared to 2.

Since tends to 0 I would be considering

Which, of course, does diverge

latex fail - not sure why

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**davros**)

How do you know the terms are integers - I don't think they are!

You basically need to be comparing the growth of the numerator with that of the denominator, which should be fairly easy to do since the n^2 terms are the same so you just need to say something about 2^n compared to 2.

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#6

I read in awe of your superior mathematics skills, if only I had chosen to listen and do the work past year nine, I may not feel like such a simpleton.

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#7

(Original post by

It has been a while but I would have looked at

Since tends to 0 I would be considering

Which, of course, does diverge

latex fail - not sure why

**TenOfThem**)It has been a while but I would have looked at

Since tends to 0 I would be considering

Which, of course, does diverge

latex fail - not sure why

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**TenOfThem**)

It has been a while but I would have looked at

Since tends to 0 I would be considering

Which, of course, does diverge

latex fail - not sure why

One more..

Summation of (n/n+1)

which series convergence test is recommended? I tried ratio which gave (n^2+2n)/(n^2+2n+1) which doesn't seem to help because this gives convergence to 1 (by diving through by n^2) which means the ratio test tells us nothing.

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#9

(Original post by

One more..

Summation of (n/n+1)

which series convergence test is recommended? I tried ratio which gave (n^2+2n)/(n^2+2n+1) which doesn't seem to help because this gives convergence to 1 (by diving through by n^2) which means the ratio test tells us nothing.

**MSI_10**)One more..

Summation of (n/n+1)

which series convergence test is recommended? I tried ratio which gave (n^2+2n)/(n^2+2n+1) which doesn't seem to help because this gives convergence to 1 (by diving through by n^2) which means the ratio test tells us nothing.

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#10

**MSI_10**)

One more..

Summation of (n/n+1)

which series convergence test is recommended? I tried ratio which gave (n^2+2n)/(n^2+2n+1) which doesn't seem to help because this gives convergence to 1 (by diving through by n^2) which means the ratio test tells us nothing.

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#11

Apart from the issue of it not necessarily being an integer (take ), you should justify that it is unbounded (the whole integer / non-negative part is entirely unnecessary, since every convergent sequence of real numbers is bounded). Try the comparison test.

The null sequence test gives it immediately.

(Original post by

One more..

**MSI_10**)One more..

(Original post by

Summation of (n/n+1)

which series convergence test is recommended? I tried ratio which gave (n^2+2n)/(n^2+2n+1) which doesn't seem to help because this gives convergence to 1 (by diving through by n^2) which means the ratio test tells us nothing.

**MSI_10**)Summation of (n/n+1)

which series convergence test is recommended? I tried ratio which gave (n^2+2n)/(n^2+2n+1) which doesn't seem to help because this gives convergence to 1 (by diving through by n^2) which means the ratio test tells us nothing.

The null sequence test gives it immediately.

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(Original post by

Apart from the issue of it not necessarily being an integer (take ), you should justify that it is unbounded (the whole integer / non-negative part is entirely unnecessary, since every convergent sequence of real numbers is bounded). Try the comparison test.

The null sequence test gives it immediately.

**BlueSam3**)Apart from the issue of it not necessarily being an integer (take ), you should justify that it is unbounded (the whole integer / non-negative part is entirely unnecessary, since every convergent sequence of real numbers is bounded). Try the comparison test.

The null sequence test gives it immediately.

Thanks for the extra verification.

Last one for today:

Summation of an=n/(n^3-1) from n=2 to n--->infinity

By comparison test, we get that an is less than 1/n^2. Since 1/n^2 converges to 0 as n--->infiniry, we conclude that an tends to 0 as n---> infinity. since 0<1, we conclude that summation of an converges

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#13

(Original post by

Thanks for the extra verification.

Last one for today:

Summation of an=n/(n^3-1) from n=2 to n--->infinity

By comparison test, we get that an is less than 1/n^2. Since 1/n^2 converges to 0 as n--->infiniry, we conclude that an tends to 0 as n---> infinity. since 0<1, we conclude that summation of an converges

**MSI_10**)Thanks for the extra verification.

Last one for today:

Summation of an=n/(n^3-1) from n=2 to n--->infinity

By comparison test, we get that an is less than 1/n^2. Since 1/n^2 converges to 0 as n--->infiniry, we conclude that an tends to 0 as n---> infinity. since 0<1, we conclude that summation of an converges

It't not the fact that 1/n^2 ->0 that is relevant here, it's the fact that converges that is important.

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#14

**MSI_10**)

Thanks for the extra verification.

Last one for today:

Summation of an=n/(n^3-1) from n=2 to n--->infinity

By comparison test, we get that an is less than 1/n^2. Since 1/n^2 converges to 0 as n--->infiniry, we conclude that an tends to 0 as n---> infinity. since 0<1, we conclude that summation of an converges

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#15

(Original post by

Summation of an=n/(n^3-1) from n=2 to n--->infinity

By comparison test, we get that an is less than 1/n^2.

**MSI_10**)Summation of an=n/(n^3-1) from n=2 to n--->infinity

By comparison test, we get that an is less than 1/n^2.

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