Nested Sigma Noation

Just saw this in that other Rubiks Cube thread, and I have no idea how to do this....

Simplified version from what they were doing:


How do you find the sum of this? Or prove that it converges / diverges? What happens when you add a third? A fourth...?

I hope you've seen these before; they're the basic definitions of integrals. In the second one, I have nested an integral within an integral, thus producing what's known as a double integral. The alternate integral notation reveals the answer to two of your three questions:

"How do you find the sum of this?"
"What happens when you add a third? A fourth...?"

Just like in a double-integral, work your way outward from the innermost integral. The exact same process occurs here with your nested sums:





The convergence/divergence tests are essentially the same as those for single summations. If one of the series produced diverges, the nested integral will diverge. For example, since your series includes a p-series of p=1, the series diverges. Alternatively, you can switch the order:





You can easily see that the series of 1's sums up to infinity so the series diverges. Unfortunately, I haven't really worked with summations so I can't tell you if there exists series that might diverge despite being composed of individually convergent series. I'm fairly certain that there exist such series, but my knowledge of this subject ends about here. I'm sure someone more qualified in this field such as Noah or Russt can further help you out.

You can easily see that the series of 1's sums up to infinity so the series diverges. Unfortunately, I haven't really worked with summations so I can't tell you if there exists series that might diverge despite being composed of individually convergent series. I'm fairly certain that there exist such series, but my knowledge of this subject ends about here. I'm sure someone more qualified in this field such as Noah or Russt can further help you out.

Need to think before i post.
When working with summations, terms can be regrouped. If a series is composed of several convergent series, regroup into the convergent parts ->Individual convergent series can be summed -> sum individual series -> whole series still converges.

You need a divergent component to make a larger compilation divergent.

I haven't had calculus beyond High School BC, so I have only seen double Integral notation - never had to deal with it.

So you would just evaluate inner most sigma (or integral) and work outwards? eg;


No idea how to find the sum of a series like that....

Or did I do that wrong.... This looks complicated and easy-to-err, but not 'hard' or 'new'. Just reapplying things I've already learned in a new setting.

I will note that the Riemann sum is only defined for the limit of deltax->0. The sum is not necessarily the same as the integral, and hence the integral test is not appropriate everywhere even though it's a very tempting test since it's so easy.

*I absolutely hate analysis.*

^That's fine, since that was a GP with regular interval.

Hadriel

So you would just evaluate inner most sigma (or integral) and work outwards? eg;

Actually, the formula is:



Take a look yourself to confirm. Thus:



The first term (1/0) should already tell you that this series is divergent. Or you can say that since this is a p-series of 1, it is divergent. Or you could do it the easier way and expand it the same way I have done in the second post:





Then the divergence becomes very obvious.

I haven't had calculus beyond High School BC, so I have only seen double Integral notation - never had to deal with it.

Here is an example of a double-integral.



Work your way outward. As you can imagine, triple integrals work the same way. The geometric interpretation of this is that the function within the integral describes the density at every point, and the integration returns the total mass of the domain, or region of the integration.

Need to think before i post.
When working with summations, terms can be regrouped. If a series is composed of several convergent series, regroup into the convergent parts ->Individual convergent series can be summed -> sum individual series -> whole series still converges.

You need a divergent component to make a larger compilation divergent.

I'm pretty sure that an infinite series of convergent series can diverge, and this time, I even have an example:



This is clearly a divergent series, even though the individual terms converge:

Just saw this in that other Rubiks Cube thread, and I have no idea how to do this....

Simplified version from what they were doing:


How do you find the sum of this? Or prove that it converges / diverges? What happens when you add a third? A fourth...?

There are some neat tricks here, which are handy whenever dealing with summations.
- As long as multiplications are only constants (relative to the summation), we can put it out of the summation, or in in the summation.
- If there are two or more summation signs "in a row", any permutation of their order will be equal to the original summation.
- You can split summations in two or more, and merge two or more summations as long as they equal the same sum.

So, even though people have shown that the series is clearly divergent, let me show you another way of showing that the series is divergent. Using the middle trick I mentioned, we can see that:



By splitting it up:



And we know that



So, clearly, this must diverge.

Also, whether individual terms converge or not does not matter: If the individual terms does in fact converge, express them as their value and not their sum. It should most likely help out. Ex. of individual converging series:

 Spoiler

The original series is only two replacements from converging though, as



converge. It's interesting to note that the series equal to 1 - you should try to prove it by using the examples mentioned earlier!

Noah

I'm pretty sure that an infinite series of convergent series can diverge, and this time, I even have an example:



This is clearly a divergent series, even though the individual terms converge:

Edit: gah, noah beat me to it

Starting with m=2 takes out only 1 term from the harmonic subseries. You eliminate the (more obvious) series of 1s, but not 1/2 + 1/3 + 1/4 + ...
Starting both at 2 eliminates all divergent components. When you do that double sum, it = 1 (from a telescoping series). Group terms like geometric series. Don't forget the geometric sum starts at 0, not 2, so you need to subtract those back out.
Or, if you like, you get sum(1/(n^2-n))

So I've proven your example wrong but haven't proven my statement right. meh. Doesn't really get us anywhere.

So I've proven your example wrong but haven't proven my statement right. meh. Doesn't really get us anywhere.

I don't see how my example is wrong. You simply offered a new example.

So, even though people have shown that the series is clearly divergent, let me show you another way of showing that the series is divergent.

I did exactly what you did in the first post, except in a less formal way.

The example you used still has a divergent component, it isn't solely composed of convergent series.

^True. The example was flawed.

Hadriel

Actually, the formula is:

Do be careful about jumping straight to a/(1-r) for something that looks like a geometric series.

Otherwise

1 +2 +4 +8 +...

Adds up to

1/(1-2) = -1

Pretty sure it doesn't add up to negative one.

the whole being able to add up to a/1-r only holds for |r| <1. So using it for when your n =1 (hence r=1) isn't valid. Not to say the rest of what you're saying is wrong. The sum of 1's definitely diverges... you just shouldn't do it by changing a geometric series to a/1-r when |r| isn't always <1.

Granted, it's quite obvious it only fails when your n=1. But I'd just like to point out... be careful.

Just noting that one can always use limits to find the value when n approachs 1. Obviously you're not going to be able to find it exactly because it doesn't exist (infinity).

^Ehh... that GP you used is for |r|< 1 i.e. infinite GP series, so why are you plonking the interval=2 example into the |r| < 1 GP formula? Nobody does a sum to infinity for r > 1... But of course your point still stands.

Nevermind. Dumping 1 into the infinite GP formula still gives you infinity, which shows that it diverges (very loosely speaking).

Hadriel

I did that because technically, it's done in the example above me. He claimed: "The first term (1/0) should already tell you that this series is divergent".

But that's saying that

1+1^2+1^3+.... = 1/(1-1)

Which isn't something you're supposed to do since a/(1-r) is only supposed to be done when |r| <1. I make this easier to understand by showing an example (of a=1, r=2) where positive, and growing, terms somehow add up to a negative number. It's all to show that you aren't supposed to do it.

Even without splitting up the double-sum of the original post the way Noah has done to avoid plugging in 1 into the geometric series formula 1/(n-1), I would still find that 1/(1-1) = 1/0 = infinity, so the double-sum in the original post diverges. Anyway:

While the sum of a geometric series formula was designed for |r|<1, 1 is so close to 0.999, 0.9999, etc. that informally, a limit can be taken to "fit" 1 into the formula.

1+1^2+1^3+ ... ≈ 0.9999^1+0.9999^2+0.9999^3+...
1/(1-1) ≈ 0.9999/(1-0.9999)

Thus, if:
(1/n)^1+(1/n)^2+(1/n)^3+... = (1/n)/(1-(1/n)) = 1/(n-1)

Then taking the limit as 1/n approaches 1-, or similarly, as n approaches 1+, results in positive infinity, which is the expected result even though the form is technically undefined.

It's a good thing I'm not taking Math B next year, else my supervisor will suck my blood for saying all that.

I will absolutely refuse to do any analysis question for my Math paper next month. I'd rather do calculus.

Hadriel

It's a good thing I'm not taking Math B next year, else my supervisor will suck my blood for saying all that.

Why so?

I will absolutely refuse to do any analysis question for my Math paper next month. I'd rather do calculus.

What's wrong with analysis? I hope to learn complex analysis some day. It seems useful.

Why so?



What's wrong with analysis? I hope to learn complex analysis some day. It seems useful.

Personal preference, it sounds like.

Analysis shouldn't be wishy-washy, like what I said just now about dumping 1 in without saying anything about taking limits... blah blah. Too anal for me.

I'm a chemist/molecular biologist, so I don't exactly care about analysis very much. Matrices and calculus are a yes-yes. Pure math is a big no-no to me. Anyway solving DEs and crunching calculus is therapeutic for the brain, since quite a bit of year 1 math doesn't need much brain cells.

Hadriel

What's wrong with analysis? I hope to learn complex analysis some day. It seems useful.

Complex analysis is not real analysis. Complex is actually fun, its just another class where you manipulate numbers and variables and see cool things. You don't really deal with super technical proofs because it's kind of assumed you already sucked at that in real analysis, and it's kept a little easier.

Real analysis is mostly proofs. Hard proofs. Extremely technical. Just for the sake of being hard. Most of the homeworks were silly things like "Prove this theorem", and you're like "but that theorem is obvious!" and then the proof turns out to be really subtle and you only get half credit for writing it poorly. Without an amazing teacher, the class inevitably sucks. No teacher at my school is amazing at teaching it.