Wouldn't the new design just get shot at differently?

Movie looks the best.

Get the regular pro unless you have big hands.

Ah, yes, how physics departments do math.

No I need to prove the cauchy condensation theorem/cuachy criterion/2ⁿ test as well.

All that just for Jaynes to fix it.

I don't care about equivalence in difficulty, I am trying to find the method for finding the slowest growing positive decreasing sequence of partial sums which is unbounded. Aka the slowest divergent non-negative series. That value seems to be S 1/n. S 1/n^p where p>1 converges.

The reason 1/n does not converge is because it goes to 0 slower than 2^p increases. This is what I wanted to find out, and from what I've seen the only way to find that out is to use 2^p (2a_n^p) without using improper integrals. Improper integrals also seem to be less general.

When a_n is 1/n^p , then 2^p (2a_n^p) = [2(1-p)]^n. Since [2^1-p]ⁿ is a geometric series, that converges when 2^1-p < 1 and diverges when 2^1-p >\ 1. Thus 1/n^p converges when 1-p < 0 <=> p > 1.

This is helpful because the infinite limit of 1/n = 0, so we need another way to determine whether or not partial sums of it converge. The 2ⁿ test produces a closed form geometric sequence of partial sums and we know both that it is an upper bound for the partial sums of 1/n and how to test it for convergence. Its convergence then implies the convergence 1/n^p , and its divergence then implies the divergence of 1/n^p -- even without knowing the closed form limit of the partial sums of 1/n.

I found a better explamation in a different book:

Sunlight too...

I noticed that Protestant democrats are the worst racists. Manifest Destiny types.

A thick hoodie and rain is not bad.

And it's not even true. People who like someone more than anyone else might get nervous and be too scared to do anything.

It's all the shit food and drugs.

That dissapoints me. It clearly had a downgraded display. I expect all pixel A series devices to be as good as the 4a 5g.

stock market... but a gift is still nice.

I'll say one thing: errections in pants are painful. I have wondered for a while why pants and shorts don't have protuding "penis pouches" that extend out of the pants to allow errections without pain.

On the next page they say that from this lemma, if S_N is bounded, then S_2^K is bounded and therefore T_K is bounded. Conversly if T_K is bounded, then S_2^K+1-1 is bounded and (by induction) 2^K+1-1 >\ K + 1, so S_N is bounded.

It's hard to follow.

No, the paper says the criterion is used to test whether a positive and decreasing series converges... the proof says the task is to show thst S_N is bounded iff T_k is bounded. So I asked how to show T_k is bounded, which is the task. The series S (a_n), with partial sums S_N, converges only if S (2^k(a_2^k), with partial sums T_K, converges. The proof is to show S (2^k(a_2^k)) converges, and thus to prove that T_K is bounded.

To do this, they use a lemma: for any natural number K, S_2^K+1-1 <\ T_K <\ 2S_2^k . They prove this lemma by using induction on K, which is what I'm struggling to follow. The initial case of K=0 is simple: S_1 <\ T_0 <\ 2S_1 <=> a_1 <\ a_1 <\ 2a_1 , since a_1 is non-negative.

Assuming K and showing it holds for K+1 is where I get lost.

Perhaps I should show you the next page:

Needs to be an infinite sequence. Here is a photo of a proof, I simply can't follow it.

Yeah, that's like the indefinite integral test. However, I'd like to use the cauchy criterion on the harmonic series.

I love this.

Yeah, all I said about (a_n) is that it is all positive and decreasing. I'm not even talking about the sum of the terms of (a_n), I'm just talking about a sequence. The sum of the terms of (a_n) only converges if the sum of 2ⁿ a_2ⁿ converges

However, to be justified in saying that I must first prove that to be true by finding out when 2ⁿ a_2ⁿ is bounded and when it isn't. It is this part that I am struggling with, because I am finding it not so easy to compare the rates of growth between 2ⁿ and different possibls (a_n)'s.

Well yeah, but I wanted to go the other direction where I use the Cauchy criterion to find out that any p>1 causes convergence. Others have said a ratio test. I haven't tried that yet.

I am given no task. I am trying to prove and understand why that exact result is true, in order to see why 1/n is the cut off for divergence.

Right now I only know that it happens, but I don't know why.