These things grow exponentially.

For any N (say N=10^29 as in your example), it is not hard to see that the number of items required is between 2^N and 2^(2N) - and that is just for the FIRST one.

I have no idea how or if it's possible to calculate exact values.

Thank you very much dear Mr. Luís Sequeira.

1, can you prove your work that the number of Un--->0 items required for the FIRST 100000000000000000000000000000 is between 2^N and 2^(2N).

2, I wish to see someone is able to design the proofs for the second and the third one and change Harmonic Series into an infinite series with infinite items each bigger than any positive constants (such as 100000000000000000000000000000).

3, If we are unable to get just three 100000000000000000000000000000, how can we get infinite 100000000000000000000000000000 from Un--->0 items in Harmonic Series -------not for mathematics but for fun or joking?

Thank you again!

Yours sincerely,

Geng

Let me show 

Let S_N = 1 + 1/2 + ... + 1/N

The standard proof of the divergence of the Harmonic series goes by showing that

S_{2N} - S_{N} ≥ 1/2:

S_{2N} - S_{N} = 1/(N+1) + ... + 1/(2N) ≥ 1/(2N) + ... 1/(2N) = 1/2

One just as easily shows that S_{2N} - S_{N} ≤ 1:

S_{2N} - S_{N} = 1/(N+1) + ... + 1/(2N) ≤ 1/(N) + ... 1/(N) = 1

from this it a simple induction to show that, starting with N=3, we always have S_{2^N} < N.

S_{2^3} = 761/280 < 3

and by induction if S_{2^N} < N, then S_{2^(N+1)} ≤ S_{2^N} + 1 < N + 1

So you need more than 2^N terms to reach a sum of N. 

Thank you very much dear Mr. Luís Sequeira.

How many is “more than” in your conclusion of “So you need more than 2^N terms to reach a sum of N. ”?

Yours sincerely,

Geng

I have no specific answer.

What I could give you is a lower bound and an upper bound. 

I agree that there is a huge gap. 

For your N=10^29 those bounds say that the number of terms of the Harmonic series you need to sum in order to reach a total of N is bigger than 2^N.

The size of this is mind-blowing. Already N is several orders or magnitude larger than, for instance, the Avogadro constant. Now picture, if you can, the size of 2 to the power N. You can't picture it... there are not enough atoms in the universe... 

You can try a few small examples of N and see what comes up, but I don't see any discernible pattern.

Denote by S(n) := 1 + 1/2 + ... + 1/n

and let H(N) be the first n such that S(n) ≥ N.

Here are few examples (computed in Python using Sympy to do exact arithmetic with rational numbers):

N=3  -  the bounds I gave you say that    8 ≤ H(3) ≤ 64

Actual value: H(3) = 11

N=4   ( 16 ≤ H(4) ≤ 256 )

Actual value: H(4) = 31

N=8 (256 ≤ H(8) ≤ 16384)

Actual value: H(8) = 1674

N=10 (1024 ≤ H(10) ≤ 1048576)

Actual value: H(10) = 12367

This one already took several minutes to compute, so I won't go any further; there is likely to be no easy formula that would get you H(N) for some N. 

Thank you very much dear Mr. Luís Sequeira.

If we are unable (no easy) to get just the first one, how can we get the second, the third, the fourth, …and change Harmonic Series into an infinite series with infinite items each bigger than any positive constants (such as 100000000000000000000000000000)?

What do you think when some people suggested that we can do it from the definition of the Riemann sum ∑1=2n+11i

Using integrals is a great idea.

I should have thought about it :-)

I have trouble following your notation, though. It turns out that we can really get a good result this way. I will write it down and post a file, since it is not easy to type mathematics in here (I don't if we can type LaTeX code directly, so it's better for me to do it separately).

Ok, now I can offer a better answer (see the attached pdf for the explanation).

The previous bounds of 2^N < H(N) < 2^{2N}

can be substantially tightened to:

                e^{N-1} < H(N) < e^N

Your idea of using the integral was key to this.

One could maybe push this idea forward and get something for your second and third steps, but I am not seeing any easy way to do this - note that even though the new estimates are much better, there is still a huge interval; I still think that reaching any sort of "formula" to calculate this given some N is extremely difficult, probably impossible.

Thank you very much dear Mr. Luís Sequeira.

1，        do you mean you can manage to get just (only) one  100000000000000000000000000000 or many more 100000000000000000000000000000 from Un--->0 items in Harmonic Series by both ways?

2，        If we can only manage to get just (only) one but unable to get three 100000000000000000000000000000, how can we get many (infinite) 100000000000000000000000000000 from Un--->0 items in Harmonic Series -------is it not for mathematics but for fun or joking at tea time?

Yours sincerely,

Geng

Let us call N=100000000000000000000000000000.

Since the sum of the harmonic series is infinite, it will reach N, 2N, 3N, etc eventually.

It is just that each time it will take more and more terms to do so as it progresses.

It will take at least e^(N-1) terms for the sum to reach N. Then it will take a lot more than that to sum another N (i.e. for the total sum to reach 2N), and so on.

You want a value for this. I can't provide one, but from the nature of infinity it is clear that the sum will reach above any real number you want. 

Thank you very much dear Mr. Luís Sequeira.

1， with the law of “the sum of the harmonic series is infinite”, people prove “it will reach N, 2N, 3N, etc eventually”.

2,  with “it will reach N, 2N, 3N, etc eventually” people prove ”the sum of the harmonic series is infinite”.

Yours sincerely,

Geng

The two statements

a) "the sum of the series is infinite" 

and

b) "the partial sums of the series will reach eventually N, 2N, 3N,... kN for a given N and for any positive integer k"

are equivalent, for any series of positive numbers.

That simply comes from the definition of the limit being +infinity.

So of course, if you assume a) you can prove b) and if you assume b) you can prove a).

It turns out that since the harmonic series is divergent, then both statements a) and b) are true when applied to the harmonic series.

Thank you very much dear Mr. Luís Sequeira.

1， with the statement of “the sum of the harmonic series is infinite”, people prove “it will reach N, 2N, 3N ,... kN eventually for a given N”.

2，  with the statement of “it will reach N, 2N, 3N ,... kN eventually for a given N” people prove ”the sum of the harmonic series is infinite”.

I think this is a logic circulation------can prove anything we wish.

Yours sincerely,

Geng

As the members from the same paradox family generated by the same classical “potential infinite--actual infinite” based science and mathematics theory system, the role of Great Achilles in Zeno’s Achilles--Turtle Race Paradox 2500 yeas ago is exactly the same thing by the title of “Brackets-Placing Artifact” good at manufacturing infinite items each bigger than 1/2, or 100, or 1000000, or 100000000000000000000000000000,…from Un--->0 items in Harmonic Series. We have to admit a truth: The conclusion of “infinite numbers each bigger than any positive constants can be produced from the Un--->0 items in Harmonic Series by brackets-placing rule based on modern limit theory " has been confirmed as a truth and a unimpeachable basic theory in our science (mathematics) means the suspended Ancient Zeno’s challenging statement of “not matter how fast Achilles runs, he will never catch up with the slowly crawling Turtle in Achilles--Turtle Race” has been confirmed with exactly the same operations and same ideas basing on modern limit theory as a “strict mathematical proven” truth and a unimpeachable theorem------it would be Great Zeno’s Achilles--Turtle Race Theorem but not Suspended Zeno’s Achilles--Turtle Race Paradox!

Five of my published papers have been up loaded onto RG to answer such questions:

1，On the Quantitative Cognitions to “Infinite Things” (I)

2，On the Quantitative Cognitions to “Infinite Things” (II)

3，On the Quantitative Cognitions to “Infinite Things” (III)

4 On the Quantitative Cognitions to “Infinite Things” (IV)

5 On the Quantitative Cognitions to “Infinite Things” (V)