Determine, using the limit comparison test, whether the series with terms (n^4 + 3n)/(5n^3 + 9) converges or diverges. Which series did you use for the limit comparison test?. I have no clue where to begin. Thanks
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Well, I can say right off the bat that this series diverges. It fails the divergence test, specifically the terms do not tend to 0. Intuitively, the numerator grows faster than the denominator, so it can't tend to 0. I could present a more precise argument, but you're very specific about using the limit comparison test, so we'll use it. The point is first to determine whether we want to prove it convergent using a convergent upper bound, or prove it divergent using a divergent lower bound. Now, it's clear we need a divergent lower bound.
The aim of the game is to play with this expression until we reach a series we know diverges, but there are rules about what we can play with. We want to make the expression smaller, so we are allowed to add positive things to the denominator, and subtract positive things from the numerator (or do other more creative things, but let's just stick to these rules for now). Both of these only make the fraction smaller, but we need to make sure we don't make it so small that it converges.
Anyway, the way I did it was to factorise the top, and notice that the factor n^3 + 3 looked a lot like 5n^3 + 9. If we did some manipulation to the denominator, we might be able to make them cancel. Specifically, if we add the positive amount 6 to the denominator, we obtain:
(n^4 + 3n) / (5n^3 + 15) = n(n^3 + 3) / 5(n^3 + 3) = n / 5
which is a known divergent series. Since the only difference is the denominator is a little larger for all n, it means this series is strictly smaller for all n. That is:
(n^4 + 3n) / (5n^3 + 9) > (n^4 + 3n) / (5n^3 + 15) = n / 5
But n / 5 diverges, so by the limit comparison test, using n / 5, the series given is divergent.
The aim of the game is to play with this expression until we reach a series we know diverges, but there are rules about what we can play with. We want to make the expression smaller, so we are allowed to add positive things to the denominator, and subtract positive things from the numerator (or do other more creative things, but let's just stick to these rules for now). Both of these only make the fraction smaller, but we need to make sure we don't make it so small that it converges.
Anyway, the way I did it was to factorise the top, and notice that the factor n^3 + 3 looked a lot like 5n^3 + 9. If we did some manipulation to the denominator, we might be able to make them cancel. Specifically, if we add the positive amount 6 to the denominator, we obtain:
(n^4 + 3n) / (5n^3 + 15) = n(n^3 + 3) / 5(n^3 + 3) = n / 5
which is a known divergent series. Since the only difference is the denominator is a little larger for all n, it means this series is strictly smaller for all n. That is:
(n^4 + 3n) / (5n^3 + 9) > (n^4 + 3n) / (5n^3 + 15) = n / 5
But n / 5 diverges, so by the limit comparison test, using n / 5, the series given is divergent.
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Compare to 1/n^2.
if lim ((n^4 + 3n)/(5n^3 + 9)) / (1 / n^2) = p, then your series must converge (since we know that 1/n^2 converges).
thus:
lim ((n^4 + 3n)/(5n^3 + 9)) / (1 / n^2)
= lim [ (n^4+3n)*n^2 ] / [ 5n^3 + 9 ]
= lim [ n^6 + 3n^3 ] / [ 5n^3 + 9]
= lim [ 1 + 3/n^3 ] / [ 5 / n^3 + 9 / n^6 ] -> divided through by n^6.
= infinity -> this reduces to (1+0) / (something that goes to 0) = 1 / (something small) = infinity.
since infinity is not bounded, we know that your series diverges.
if lim ((n^4 + 3n)/(5n^3 + 9)) / (1 / n^2) = p, then your series must converge (since we know that 1/n^2 converges).
thus:
lim ((n^4 + 3n)/(5n^3 + 9)) / (1 / n^2)
= lim [ (n^4+3n)*n^2 ] / [ 5n^3 + 9 ]
= lim [ n^6 + 3n^3 ] / [ 5n^3 + 9]
= lim [ 1 + 3/n^3 ] / [ 5 / n^3 + 9 / n^6 ] -> divided through by n^6.
= infinity -> this reduces to (1+0) / (something that goes to 0) = 1 / (something small) = infinity.
since infinity is not bounded, we know that your series diverges.