| Posted on Wednesday, 22 October, 2003 - 06:18 pm: |
Hi! Here are two lovely questions, who can solve them? 1. 1+(1/2)+(1/3)+...+(1/n)=? where n does not tend to infinity. 2. Is it a way to prove 1+(1/2)2+(1/3)2+... = (p2)/6, where ... means tending to infinity.
| Posted on Wednesday, 22 October, 2003 - 06:26 pm: |
1. There is the approximation that the sum is approximatly (ln n) + 1/2, although I don't know how to derive this or whether there is an exact answer.
| Posted on Wednesday, 22 October, 2003 - 07:03 pm: |
For (2), if you are familiar with Fourier series, it is easy. If not, there is a way to do it from some trigonometric identities, but you need to be familiar with complex numbers, manipulating roots of polynomials etc.
Andre
| Posted on Wednesday, 22 October, 2003 - 09:12 pm: |
For (1), I think we can in fact derive it to obtain a general expression of the sum of this sequence. And this expression should have a property is that when n tends to infinity, the outcome will be infinity as well.
But I can't think out how to do it. However, I don't think (In n) + 1/2 will work. Since if n = 1 in the sequence, the outcome will be 1, but from (In 1) + 1/2, it's 1/2. (eg.for n = 2, the answer from sequence is 3/2, but (In n) + 1/2 is an approximation.) What do you think?
| Posted on Wednesday, 22 October, 2003 - 09:30 pm: |
Sorry Andre, I'm confused.
And what is Fourier series by the way?
| Posted on Wednesday, 22 October, 2003 - 09:47 pm: |
David, the approximation I mentioned earlier is valid for large n.
| Posted on Wednesday, 22 October, 2003 - 10:36 pm: |
Chris, there is a better approximation: it is ln(n)+g for large values of n. The number g is known as Euler's constant which is 0.5772156649...
| Posted on Thursday, 23 October, 2003 - 11:40 am: |
David,
Fourier series are something you will meet at university. The idea, though, is straightforward.
Suppose we have a periodic function [that is, f(x)=f(x+k) for any x, where k is the period]. Then this function can be represented as the sums of sin's and cos's.
If you are familiar with Excel, you can best get the idea with a few examples. In each case, plot the function from x=0 to 31 (with spacing of, say, 0.2)
Eg 1:
a) sin(x)
b) sin(x) + (1/3)*sin(3x)
c) sin(x) + (1/3)*sin(3x) + (1/5)*sin(5x)
d) sin(x) + (1/3)*sin(3x) + (1/5)*sin(5x) + (1/7)*sin(7x)
etc
can you guess what this is converging to?
Eg 2:
e) sin(x)
f) sin(x) + (1/2)*sin(2x)
g) sin(x) + (1/2)*sin(2x) + (1/3)*sin(3x)
h) sin(x) + (1/2)*sin(2x) + (1/3)*sin(3x) + (1/5)*sin(5x)
etc
can you guess what this is converging to?
It is bizarre that things as smooth as sine waves can add up to functions that are not continuous!
Now it turns out that if you give me the original function [f(x)], it is fairly easy to compute the coefficients of sin and cos (you need to do an integration). Note also, that if a function requires cosines only, then the value of the original function at x=0 must equal the sum of the coefficients of the cosines [because all the cos's evaluate to 1 at x=0].
In your case, if my memory serves me correctly, you need to pick a function that is a repeating parabola shape, and the coefficients of the cosines are those of your original sum [actually, if I had to do it, I would use a simpler function and something called Parseval's theorem instead -it's easier!]
Andre
| Posted on Thursday, 23 October, 2003 - 02:15 pm: |
For 1, i went about it in several ways, but none of them managed to achieve anything in particular.
Firstly i found the ratios between each term which were 1/2, 2/3, 3/4.... so i saw this was a difference of (n-1)/n, and so an =(n-1)!/n!
However this ratio is changing and so it cannot be plugged into a formula.
Next i rewrote 1/2 + 1/3 + 1/4 ....
As a single fraction of denominator 1x2x3x4....=n!
So
, however this cant be
summed any easier than the original version since n!/2
etc must be calculated individually.Thirdly i looked at the power series of ln.
This can be found as
ln(1+x)=x-(x^2)/(2) + (x^3)/(3) - (x^4)/(4) + (x^5)/(5)....
So if we substitute x=1, we get something that looks similar to what we want. ln(1+1)=ln(2)
Now if we take our original series and minus ln(2) we get 0+1+0+(1/2)+0+(1/3) minus whatever is left of ln(2).
Therefore for large values of n, Sn -ln(2)=Sn/2 . Rearranging giving:
Sn =Sn/2 +ln(2)
Maybe this is what Chris meant?
| Posted on Thursday, 23 October, 2003 - 03:24 pm: |
This completely non-rigorous yet rather interesting :) method (or something similar) is due to Euler for (2): Consider the function f(x)=(sinÖx)/Öx It is easy to verify that the roots of this are p2, 4p4, 9p2,... Also, the series expansion of f is, f(x)=1-x/3!+x2/5!-... [Here's where it gets unrigorous...] Assume (as would be the case if there was a finite number of roots) that f(x) can be written as (1-x/p2)(1-x/4p2)(1-x/9p2)... (i.e. as an infinite product of brackets of the form (1-x/R) where R are the roots of f) Now, from this product we can see that the coefficient of x would be -(1/p2+1/4p2+ 1/9p2+...) Comparing this with the coefficient of x in the series expansion gives: 1/p2+1/4p2+1/9p2+...=1/6 Multiplying through by p2 gives the required result... Marcos
| Posted on Thursday, 23 October, 2003 - 03:35 pm: |
James, As Mark has said, ln(n)+g is better than Chris's approximation but as g is about 1/2 Chris's approximation is just a simplification of Mark's. Mark's approximation follows (although it's kinda cheating :) ) from the definition of g as the limit of [1+1/2+1/3+...+1/n-ln(n)] as n®¥ Marcos
| Posted on Thursday, 23 October, 2003 - 05:11 pm: |
...once you have proved the limit exists of course.
| Posted on Thursday, 23 October, 2003 - 05:42 pm: |
Yeah (but don't ask me how!
)
| Posted on Thursday, 23 October, 2003 - 05:55 pm: |
(Leafs through bookshelf) One nice approach is to define an=log(1+1/n)-1/(n+1), and show that 0 < an < 1/(2n(n-1)). Thus
|
¥ å n=1 | an |
converges, and its limit turns out to be 1-g. Alternatively one can do various tricks with integrals. David
| Posted on Friday, 24 October, 2003 - 10:39 am: |
A very good (the best?) reference for questions similar to (1) is Whittaker + Watson. For (2), you can find more than a dozen proofs here My favourite is using the series for cot x (proof 7). It can be easily expanded to find z(2m) Demetres
| Posted on Wednesday, 05 November, 2003 - 01:10 pm: |
In answer to the first:
Taemun