How to Prove Convergence

[upper flax]

Hello so I'm wondering if anyone has a quicker way of proving convergence of my function
a_n = \frac{1}{\sqrt[3]{n^4 - \frac{2}{3}}}

What I found out was the series b_n = n^(-4/3) converges by p-test thing -4/3 < -1 and I found that c_n = b_(n-1) must also then converge by tails
Since i'm trying to do it by comparison I found the original isn't less than b_n which is why I found c_n because the original is less than c_n and I proved this by rearranging and showing that for all n >= 1, the
1/(cn)^3 = (n^4 - g(x)) and I showed that g(1) = -1, and g'(x) < 0 implying decreasing which means g(x) x> 1 is always less than -1
which implies 1/(cn)^3 <= n^4 - 1 < n^4 - 2/3
which means 1/(cn)^3 < 1/(an)^3 which means a_n < c_n for any n >= 1
and since c_n converges, a_n must by the limit comparison test.

as you can see this is such a lengthy way of doing it and I was wandering if there is aynthing quicker since the question isn't really worth many marks compared to the others:

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[upper flax]

Hello I see there is an equivalent post but I don't get what they were going on about : https://discord.com/channels/624314920158232616/1236534369548501073

[native plume]

Hey man

In all fairness I think a comparison test is plenty good already

One way to see it is the following

when n is large, crt(n^4 - 2/3) behaves like n^4/3

so with 4/3 larger than 1, the series of 1/n^(4/3) converges

and so does the series of crt(n^4 - 2/3)

but by plain comparison

to be fair the method is still similar

you can say that n^4 - 2/3 is greater than (n^4)/2 when n is large

[upper flax]

yeah I know intuitvley n^4 and it - 2/3 are the same but I'm kind of doing a course where we have to be very rigourous in what we do

[native plume]

or something

[native plume]

To be fair that is already perfectly legal

there is a theorem that says you can do that

Though I do not know what it is called in English

I can't give you the resource in English since I've never found it before

I'm referring to the first dot here

[native plume]

Again, in case of doubt, consider that n^4 - 2/3 > (n^4)/2 for large n

so clearly crt(n^4 - 2/3) > crt(n^4/2) = n^(4/3)/2^(1/3)

so 1/ crt(n^4 - 2/3) < crt(2)/n^(4/3)

that's probably a more convincing comparison in your standards

[upper flax]

yeah I just did the maths behind this and its simply n>= 2 then this expression holds
which implies
the original thing < (2 ^ 1/3) / n ^ 4/3
and since n^-4/3 converges so does any multiple of it (ie 2^1/3 * n^-4/3)

yeah

okay thank you this seems to be much easier and quicker I think I'll try this way instead of my unneccisarily complicated way

[native plume]

Well in practice it's much better to use equivalence

than try to find some fancy bounds

because as said before, that is a bona fide theorem

that incidentally gives you the rate of convergence for free

[upper flax]

well I only need to show the sequence is convergent using the limit comparison test (ie find a similar function)

but to be honest I have zero idea what I'm doing in this course and its week 11

as I said I spent all day on this one thing activley avoiding the other 70 marks of this homework sheet

idk whats wrong with me lol but thank you for helping its gave me ideas/answers

[native plume]

I don't know what exactly is demanded of you but if you just need to show convergence or divergence for a bunch of series, it's not exactly complicated most of the time

the big picture is that since you don't know how to compute most series that exist, you can still compare them to some series known to mankind

[upper flax]

these are the other ones I still have to do <
the last one I can use the ratio test of a_n+1 / a_n and show its < 1
but b) (the log one) I think I have to use the alternating series Leibniz test

[native plume]

that sounds about legitimate

[upper flax]

okay thats good

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[upper flax]

okay thanks @native plume for your help

[silent belfry]

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[upper flax]

+close