Does this convergence? Justify your answer..

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MSI_10
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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?
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MSI_10
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Anddddd epic thread title grammar fail :woo:
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davros
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(Original post by 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?
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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TenOfThem
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(Original post by 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.
It has been a while but I would have looked at

\dfrac{1+\frac{2^n}{n^2}}{1+\fra  c{2}{n^2}}

Since \frac{2}{n^2} tends to 0 I would be considering\frac{2^n}{n^2}

Which, of course, does diverge

latex fail - not sure why
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MSI_10
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(Original post by 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.
ah yeah I meant positive real numbers... it's provided in notes that exponential growth, is obviously, bigger than constant ''growth'' since constants do not change therefore 2^n>=2 for all n belonging to natural numbers.
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blu95
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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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davros
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(Original post by TenOfThem)
It has been a while but I would have looked at

\dfrac{1+\frac{2^n}{n^2}}{1+\fra  c{2}{n^2}}

Since \frac{2}{n^2} tends to 0 I would be considering\frac{2^n}{n^2}

Which, of course, does diverge

latex fail - not sure why
Yeah, I was being a bit lazy when I glanced at it earlier - the point I was trying to make was that the numerator would grow much more quickly than the denominator, so it's not sufficient just to compare to the constant term.
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MSI_10
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(Original post by TenOfThem)
It has been a while but I would have looked at

\dfrac{1+\frac{2^n}{n^2}}{1+\fra  c{2}{n^2}}

Since \frac{2}{n^2} tends to 0 I would be considering\frac{2^n}{n^2}

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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electriic_ink
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(Original post by 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.
Compare to Harmonic Series.
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davros
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(Original post by 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.
If you really are summing a series with that as the general term then it won't converge in a million years because it fails the basic test that the general term must tend to 0 as n -> infinity
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BlueSam3
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Apart from the issue of it not necessarily being an integer (take n=1), 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.

(Original post by MSI_10)
One more..
(Original post by 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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MSI_10
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(Original post by BlueSam3)
Apart from the issue of it not necessarily being an integer (take n=1), 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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davros
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(Original post by 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
NO - you need to compare the SUM of your sequence with the SUM of 1/n^2.

It't not the fact that 1/n^2 ->0 that is relevant here, it's the fact that \Sigma (1/n^2) converges that is important.
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BlueSam3
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(Original post by 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
That doesn't work. (a_n) \to 0\not\Rightarrow \sum a_n &lt; \infty (a_n = \frac{1}{n} being the canonical example). I think you're trying to apply the ratio test to establish that \sum \frac{1}{n^2} converges, but have forgotten to take ratios. It won't work anyway, since the ratios approach 1, so the ratio test is inconclusive.
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DFranklin
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(Original post by 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.
Aside from the other objections given, a_n is actually greater than 1./n^2.
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