series convergence

[final galleon]

Use the hint to rewrite the series so that the terms are of the form -sin(pi(2-√6)^k). Then use the fact that sin(x) ≤ x when x ≥ 0 to see that the negative of that series is majorized by a converging geometric series

Also note that sin(pi(2-√6)^k) will be non-negative for non-negative k

[sharp hearth]

manifesting good luck

manifesting good luck

[final galleon]

You have -sin(pi(2-√6)^k) ≥ -pi(2-√6)^k because sign changes when you negate

Yes, because for example the series 1 + 1/2 + 1/4 + ... always has larger terms than -1 - 1 - 1 - ... but the latter doesn't converge.

[final galleon]

But like you do get that the series with pi(2-√6)^k majorizes the one with sin(pi(2-√6)^k) and series converges if and only if it converges when all terms are negated

Ah right. I did make an error. I somehow thought 2 > √6 lol

Let's see how we can fix the argument

Okay well sum of two consecutive terms is positive. So we get a positive term series with the same value if we group terms in pairs

And now the argument works again

1/n approaches 1/n^2 as n gets larger, but sum 1/n diverges while with 1/n^2 it converges so I don't think argument like that works

[final galleon]

But this works. Group two consecutive terms. It doesn't change the value of the series and now geometric series with two consecutive terms also grouped majorizes this

The sum of two consecutive terms is positive because sin(x) is an increasing function when 0 ≤ x ≤ pi/2

Actually there is still a problem. Because the sign changes also in the geometric series

What we can do though is replace terms by their absolute values and show that the series converges absolutely

Which implies that the series converges

This lets us get rid of these sign problems

[final galleon]

Also this grouping isn't valid without absolute convergence or some extra argument. Otherwise we could conclude that 1 - 1 + 1 - 1 + ... converges

But we don't need to do it now anyway

[sharp hearth]

manifesting good luck

[final galleon]

Yeah

No problem

You too!

Yes

It is not enough that terms of a series approach terms of a converging series to show convergence

Eventually 1/n approach 1/n^2 i.e. 1/n-1/n^2 approaches 0 so using this argument 1/n converging is equivalent to 1/n^2 converging.

Just because difference of terms of two series converges to zero doesn't mean their convergence is tied together. One can diverge and other can converge

[final galleon]

Okay so by approaching they mean that f(n) is O(g(n)) which means that eventually f(k) ≤ Mg(k) for some constant M. Convergence of sum of g does now indeed imply the convergence of sum of f by the majorization argument we used. So with this definition of approaching the argument is correct and is pretty much equivalent to what we did here

Actually not so sure how you incorporate the sin into that. sin(pi(a+b)) = 0 so sin(pi a) isn't O(sin(pi(a+b)))

Yes. How I interprered approaching is that the limit of the difference is zero which is one way to define it