# Thread: Nth term of a harmonic series

1. Can you generalize the nth term of a harmonic series such as this one??

can anyone provide some explanation, or a link perhaps other than google hits.

thank you

2.

3. Originally Posted by Heinsbergrelatz
Can you generalize the nth term of a harmonic series such as this one??

can anyone provide some explanation, or a link perhaps other than google hits.

thank you
I think you meant this.

which is also known as or the harmonic number

It's closely related to the natural logarithmic function.

4. Im still confused about one point, like geometric progression or arithmetic progression, when they give you the first tern and the common ratio or difference, you can manage to calculate the say 600th term even, or 30 or 1000 whatever. How do you that with Harmonic series???

does harmonic series always go till infinity? making it an infinite series? can it not be like get the sum of 10 or 50 terms like G.P. and A.P.?

sorry if my understanding with harmonic series seems poor..

thank you

5. Originally Posted by Heinsbergrelatz
Im still confused about one point, like geometric progression or arithmetic progression, when they give you the first tern and the common ratio or difference, you can manage to calculate the say 600th term even, or 30 or 1000 whatever. How do you that with Harmonic series???

does harmonic series always go till infinity? making it an infinite series? can it not be like get the sum of 10 or 50 terms like G.P. and A.P.?

sorry if my understanding with harmonic series seems poor..

thank you
You don't.

Unlike the very simple geometric and arithmetic progression there is no known closed form expression for the n'th partial sum of the the harmonic series.

So far as I know it is still an open question as to the smallest number n for which the nth partial sum of the harmonic series exceeds 100.

6. Originally Posted by Heinsbergrelatz
Im still confused about one point, like geometric progression or arithmetic progression, when they give you the first tern and the common ratio or difference, you can manage to calculate the say 600th term even, or 30 or 1000 whatever. How do you that with Harmonic series???

does harmonic series always go till infinity? making it an infinite series? can it not be like get the sum of 10 or 50 terms like G.P. and A.P.?

sorry if my understanding with harmonic series seems poor..

thank you

There is a massive difference between this and G.P. and A.P.

7. i always wondered what notation was, the notation next to your sigma. What is it?(its not university mathematics, so im not quite familiar with that symbol..)

also how does one solve these kind of equations??

thank you

8. Originally Posted by Heinsbergrelatz
i always wondered what notation was, the notation next to your sigma. What is it?(its not university mathematics, so im not quite familiar with that symbol..)

also how does one solve these kind of equations??

thank you
That symbol is the sumbol for a product of the elements of a series, whereas a sigma is the symbol for the sum of the elements of a series.

Unfortunately what you see are just two expressions, neither of which admits a closed form representation, but either of which might be useful in some (at this point unknown) situation (assuming that the expressions are correct as I have not checked them)>

9. Thank for the reply, so how would one actually approach the question?
can you help me solve it? do you use the Lambert function?

well here is the actual question, not related to the equation i gave above.

-Bo-Youn and Ken are to begin a savings program. Bo-youn saves 1$int the first week, 2$ in the second and 4$in the 3rd in a G.P. Ken saves 10$ in the first week, 15 on the second 20 on the third in a A.P. When would have bo-youn saved more than Ken?

this is actually a very simple questions, where you can even solve through trial and error. But how would you solve it??

Many Thanks

10. Originally Posted by Heinsbergrelatz
Thank for the reply, so how would one actually approach the question?
can you help me solve it? do you use the Lambert function?

well here is the actual question, not related to the equation i gave above.

-Bo-Youn and Ken are to begin a savings program. Bo-youn saves 1$int the first week, 2$ in the second and 4$in the 3rd in a G.P. Ken saves 10$ in the first week, 15 on the second 20 on the third in a A.P. When would have bo-youn saved more than Ken?

this is actually a very simple questions, where you can even solve through trial and error. But how would you solve it??

Many Thanks
What connection do you see between your question and the harmonic series ? I see none. An arithmetic or geometric series has simply described partial sums. Not so with the harmonic series.

11. Actually the harmonic series was one of my doubts to the question i found in my textbook, which i didn't post up. the mixed equation of of x and 2^x was the one linked to A.P. and G.P. so i moved on to G.P and A.P. not linking it from H.P. the harmonic series i already solved in my textbook, so i had an extra inquiry that's why i asked about how to solve the x with 2^x equation-which is linked to combined G.P. and A.P.

12. Harmonic series Σ(k=1,n)1/k = ln(n) + γ + o(1), where γ is Euler constant.

13. Originally Posted by mathman
Harmonic series Σ(k=1,n)1/k = ln(n) + γ + o(1), where γ is Euler constant.
and o(1) is indeterminate and describes only the limitation on growth -- so this does not answer the question.

14. Originally Posted by DrRocket
Originally Posted by mathman
Harmonic series Σ(k=1,n)1/k = ln(n) + γ + o(1), where γ is Euler constant.
and o(1) is indeterminate and describes only the limitation on growth -- so this does not answer the question.
o(1) [not 0(1)] means the error -> 0 as n -> ∞. The answer I gave is the best that can be given.

15. Originally Posted by mathman
Originally Posted by DrRocket
Originally Posted by mathman
Harmonic series Σ(k=1,n)1/k = ln(n) + γ + o(1), where γ is Euler constant.
and o(1) is indeterminate and describes only the limitation on growth -- so this does not answer the question.
o(1) [not 0(1)] means the error -> 0 as n -> ∞. The answer I gave is the best that can be given.
o(1) and O(1) each signify a quantity with certain growth behavior at infinity. They are known as the Landau symbols.

I stated earlier that there is no known closed form expression for the nth partial sum of the harmonic series. That is quite true. Whether or no the estimate that you provided is the "best" that can be given is another matter. I am not sure what the best known result is. Are you?

If you are certain as to the best known result, a reference to the literatiure would be in order.

16. http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

See "rate of divergence" in the article.

17. Originally Posted by mathman
http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

See "rate of divergence" in the article.
All that is said in that article is the content of your earlier post, which has been known for quite a long time. That is quite a long way from an assertion that the estimate given is the best known (i.e. sharpest) for the rate of divergence. This may well be the best known result, but it is not clear to me that such is the case.

18. I have no inside knowledge as to whether or not it is the best result. However I strongly suspect that if there were a better result it would have been mentioned in the article.

19. Originally Posted by mathman
I have no inside knowledge as to whether or not it is the best result. However I strongly suspect that if there were a better result it would have been mentioned in the article.
That, or the people who know of it aren't avid users of wikipedia, meaning that it wont have any mention. Or, quite possibly, no one has come up with a better descriptor for whatever reason, which would not mean that a better one doesn't exist.

20. Originally Posted by mathman
I have no inside knowledge as to whether or not it is the best result. However I strongly suspect that if there were a better result it would have been mentioned in the article.
You have great confidence in thos who write Wikipedia articles.

I do not share that confidence.

21. Originally Posted by DrRocket
Originally Posted by mathman
I have no inside knowledge as to whether or not it is the best result. However I strongly suspect that if there were a better result it would have been mentioned in the article.
You have great confidence in thos who write Wikipedia articles.

I do not share that confidence.
I won't argue the point in general. However the particular article I referred to appears to be a thorough and accurate description of the problem.

22. Originally Posted by mathman
Originally Posted by DrRocket
Originally Posted by mathman
I have no inside knowledge as to whether or not it is the best result. However I strongly suspect that if there were a better result it would have been mentioned in the article.
You have great confidence in thos who write Wikipedia articles.

I do not share that confidence.
I won't argue the point in general. However the particular article I referred to appears to be a thorough and accurate description of the problem.
That article is accurate. But thorough is in the eye of the beholder. It aeems to contain all that most readers would want. However, the question as to whether the growth estimates are the sharpest and best known is one for real experts, and that is a completely different kettle of fish.

Definition: Expert (in a specifried field). Someone who knows everything that is known in the field, knows the important open questions, knows the leading research approaches to those questions, knows all of the researches in the area ans their graduate stucents and knows who is pursuing each research approach and their general level of progress. The knowledge in question is up to date as of five minutes ago.

23. Table of integrals series and products by I.S. Gradshteyn and I.M. Ryzhik (2007)

Above contains complete answer to original problem, i.e. error term after summing n terms. 2/n + .... (... is explicitly given).

24. Originally Posted by mathman
Table of integrals series and products by I.S. Gradshteyn and I.M. Ryzhik (2007)

Above contains complete answer to original problem, i.e. error term after summing n terms. 2/n + .... (... is explicitly given).
Formula 0.131

But you still do not understand the problem. It gives the error term, yes, but not in a form in which the decay of the error to zero is readily estimated.

Again, so far as I know the question of the smallest value of n for which the nth partial sum exceeds 100 is till an open question. Were the error term put in a form in which could be readily evaluated, this problem would then be easily solved.

There is more to understanding a problem than simply writing down a complicated expression and declaring it to be the solution. But to understand this you have to gain deeper experience with and knowledge of mathematics.

It is very easy to see that the harmonic series grows about like ln(n), but to obtain better estimates of the rate of growth (or the rate of decay of the difference between the nth partial sum and ln(n)) is a difficult problem. Apparently it is difficult to understand the very nature of the problem as well.

25. Again, so far as I know the question of the smallest value of n for which the nth partial sum exceeds 100 is till an open question. Were the error term put in a form in which could be readily evaluated, this problem would then be easily solved.
Since ln(n) + γ is a very good approx., start from n=exp(100-γ) and check nearby values.

26. Originally Posted by mathman
Again, so far as I know the question of the smallest value of n for which the nth partial sum exceeds 100 is till an open question. Were the error term put in a form in which could be readily evaluated, this problem would then be easily solved.
Since ln(n) + γ is a very good approx., start from n=exp(100-γ) and check nearby values.
Nope.

That is both naive and ineffective as a proof. You don't understand the problem.

27. I didn't know there was a problem. It sounded like a calculation. Calculate n so that sum to n is > 100 and sum to n-1 is < 100.

28. Originally Posted by mathman
I didn't know there was a problem. It sounded like a calculation. Calculate n so that sum to n is > 100 and sum to n-1 is < 100.
It's a ridiculously large number, and the harmonic series doesn't follow the natural logarithm close enough to make this a "great" approximation

29. Originally Posted by Arcane_Mathematician
Originally Posted by mathman
I didn't know there was a problem. It sounded like a calculation. Calculate n so that sum to n is > 100 and sum to n-1 is < 100.
It's a ridiculously large number, and the harmonic series doesn't follow the natural logarithm close enough to make this a "great" approximation
I am not sure what you consider "close enough". However for large n the correction to the sum = ln(n) + γ is ~ 2/n.

30. Originally Posted by mathman
I didn't know there was a problem. It sounded like a calculation. Calculate n so that sum to n is > 100 and sum to n-1 is < 100.
Nope.

It is not a calculation since you have no effective means by which to calculate. If you disagree then please provide the necessary calculation (you will find that you cannot do it in any reasonable time, even with a computer).

This simply demonstrates the thesis that you don't understand the problem.

31. First approx. n+2=e^(100-γ). This takes care of the first order correction. The formula described previously in I.S. Gradshteyn and I.M. Ryzhik can be used to get n more precisely. I will admit that the calculation would be horrendous, since n is a very large number.

32. Originally Posted by mathman
First approx. n+2=e^(100-γ). This takes care of the first order correction. The formula described previously in I.S. Gradshteyn and I.M. Ryzhik can be used to get n more precisely. I will admit that the calculation would be horrendous, since n is a very large number.
The calculation is sufficiently horrendous that it cannot be done in any reasonable time, even with a computer. The fact that there is some algorithm that would lead to a solution in a few million years is completely irrelevant. This is a trivial observation. The problem is interesting precisely because such brute force approaches do not work.

Any mathematician can see what you describe as a "first order correction"-- it is trivial and obvious. It does absolutely nothing whatever toward solving the problem of determining exactly the smalles value of n for the nth partial sum exceeds 100. Nothing.

You are talking like a physicist, not a mathematician. The formula of Gradstein and Rytzik is completely irrelevant. You do not understand the problem.

BTW be careful with Gradsteina and Rytzik. It is a very useful reference. But you need to check the solutions given. A few of them are wrong. In this case the solution is correct, but it is not helpful for this particular problem.

It is generally recommeded that when you have dug a hole for yourself you should stop digging. Your arguments are only serving to confirm that you don't understand the problem here.

33. The calculation is sufficiently horrendous that it cannot be done in any reasonable time, even with a computer.
Can you prove it?

34. Originally Posted by mathman
The calculation is sufficiently horrendous that it cannot be done in any reasonable time, even with a computer.
Can you prove it?
Yes. Can't you ?

Hint: You have nearly done it yourself, noting the approximate number involved from which you can estimate the number of arithmetic operations that need to be performed, which can then be translated into time required to perform those operations on a computer.

35. My estimate requires a modest number of arithmetic operations.

Let 10^x = e^(100-γ). x=(100-γ)/ln(10). A quick calculation gives 44 > x > 43. This means that one needs x to about 46 or 47 significant figures to get n (in the range 10^43 to 10^44) to the nearest integer. That is the horrendous part of the calculation - it shouldn't take too long on any current computer.

After the 2/n first order error estimate, further errors ~ 1/n^2.

36. Originally Posted by mathman
My estimate requires a modest number of arithmetic operations.

Let 10^x = e^(100-γ). x=(100-γ)/ln(10). A quick calculation gives 44 > x > 43. This means that one needs x to about 46 or 47 significant figures to get n (in the range 10^43 to 10^44) to the nearest integer. That is the horrendous part of the calculation - it shouldn't take too long on any current computer.

After the 2/n first order error estimate, further errors ~ 1/n^2.
This absurd.

The problem is to find the smalles value of n for which

Your statements don't even seem to apply. If you think you can solve the problem then go ahead and do it and show us. If not put a sock in it.

37. Originally Posted by DrRocket
Originally Posted by mathman
My estimate requires a modest number of arithmetic operations.

Let 10^x = e^(100-γ). x=(100-γ)/ln(10). A quick calculation gives 44 > x > 43. This means that one needs x to about 46 or 47 significant figures to get n (in the range 10^43 to 10^44) to the nearest integer. That is the horrendous part of the calculation - it shouldn't take too long on any current computer.

After the 2/n first order error estimate, further errors ~ 1/n^2.
This absurd.

The problem is to find the smalles value of n for which

Your statements don't even seem to apply. If you think you can solve the problem then go ahead and do it and show us. If not put a sock in it.
I don't have access to computer programs to make calculations to 47 significant figures. However from your comments I gather you don't understand what I am proposing.

38.

39. Originally Posted by MagiMaster

So apparently some did manage to determine that the first n for which the nth partial sum exceeds 100 is

15092688622113788323693563264538101449859497

and thus the problem has been solved. However, it is interesting to note that what is given is only the purported answer and not a proof that the claim is correct or a statement as to how the result was obtained. http://www.research.att.com/~njas/sequences/A082912

Some other interesting work related to this question.

http://www.ams.org/journals/proc/198...-0796451-8.pdf

http://arxiv.org/PS_cache/math/pdf/0402/0402354v5.pdf

http://www.ams.org/journals/mcom/197...-0440862-0.pdf

40. At the top of this entry: http://www.research.att.com/~njas/sequences/A002387 they give the formula used to calculate that and why it should give the right answer. (Which basically boils down to the difference between the lower bound and the upper bound being too small to contain two integers.)

41. Originally Posted by MagiMaster
At the top of this entry: http://www.research.att.com/~njas/sequences/A002387 they give the formula used to calculate that and why it should give the right answer. (Which basically boils down to the difference between the lower bound and the upper bound being too small to contain two integers.)
What I see is a statement that the estimates are based on a heuristic argument -- in other words not a real proof. That is a bit disconcerting. In fact, if the argument is really only heuristic then the problem remains open.

42. http://mathworld.wolfram.com/HarmonicSeries.html

has the following:
The number of terms needed for to exceed 1, 2, 3, ... are 1, 4, 11, 31, 83, 227, 616, 1674, 4550, 12367, 33617, 91380, 248397, ... (Sloane's A004080; DeTemple and Wang 1991). Using the analytic form shows that after terms, the sum is still less than 20. Furthermore, to achieve a sum greater than 100, more than 1.509x10^43 terms are needed!
(Boas and Wrench 1971; Gardner 1984, p. 167).

More generally, the number of terms needed to equal or exceed 10 and 100 are 12367 and 15092688622113788323693563264538101449859497
(Sloane's A096618).

(I had to edit it slightly - you can look up the original.)

To satisfy my own curiosity I calculated the number of terms based on the formula I described. However I could do it only to 10 significant figures - agrees with the given value.

43. Originally Posted by mathman
http://mathworld.wolfram.com/HarmonicSeries.html

has the following:
The number of terms needed for to exceed 1, 2, 3, ... are 1, 4, 11, 31, 83, 227, 616, 1674, 4550, 12367, 33617, 91380, 248397, ... (Sloane's A004080; DeTemple and Wang 1991). Using the analytic form shows that after terms, the sum is still less than 20. Furthermore, to achieve a sum greater than 100, more than 1.509x10^43 terms are needed!
(Boas and Wrench 1971; Gardner 1984, p. 167).

More generally, the number of terms needed to equal or exceed 10 and 100 are 12367 and 15092688622113788323693563264538101449859497
(Sloane's A096618).

(I had to edit it slightly - you can look up the original.)

To satisfy my own curiosity I calculated the number of terms based on the formula I described. However I could do it only to 10 significant figures - agrees with the given value.
I saw the same thing. I also saw the explanation, and the statement that a critical step is based on heuritic reasoning. All that I can conclude from this is that the harmonic number for 100 is close to the number cited, but I do not see clear evidence of a rigorous proof that that Sloane has actually proved that 15092688622113788323693563264538101449859497 is indeed H(100). The problem as stated is not to estimate H(100) but to determine it exactly. If you see a rigorous proof that they have done then I would like to see the details.

44. The actual answer could only be that or that minus 1 from what I understand of the formula, and the probability of it being that minus one decreases rapidly as n increases. But yeah, no one's actually calculated it directly, so it's still possible.

BTW, it wasn't Sloane who calculated these. T. D. Noe made the table using Hickerson's formula.

45. Originally Posted by MagiMaster
The actual answer could only be that or that minus 1 from what I understand of the formula, and the probability of it being that minus one decreases rapidly as n increases. But yeah, no one's actually calculated it directly, so it's still possible.

BTW, it wasn't Sloane who calculated these. T. D. Noe made the table using Hickerson's formula.
The problem of determining the harmonic number of 100 exactly is considerably more difficult than that of estimating it to within 1. The issue is not really finding the number, but rather finding a method for determining it exactly and proving that one is correct. As with much of mathematics it is the technique and reasoning that are of interest more so than the actual answer.

46. Computer science is the same way. It's a pretty good start though.

47. Originally Posted by MagiMaster
Computer science is the same way. It's a pretty good start though.
Sure it is a good start. The first step in mathematical research is to guess the answer or the theorem. The second step is to prove that the guess is correct.

This gives one a pretty good guess as to the answer. But it is not complete without the rigorous proof that the guess is in fact correct.

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