Summing Reciprocals

¡@¡@¡@¡@¡@¡@¡@ "The discovery of an infinite number of numbers summing to one value must be ranked as one of the most outstanding achievements of humankind. The whole idea flies in the face of intuition. How can we take an infinite of numbers, add them up, yet end up with a finite sum?"¡@¡@ --- Mathematical Mysteries

¡@¡@¡@¡@¡@¡@¡@ This is especially true when it comes to summing reciprocals of squares, as the sum contains £k2 , which seems to have absolutely nothing to do with the terms summed up.

¡@¡@¡@¡@¡@¡@¡@ When the problem first landed on the hands of Jakob Bernoulli (1654 -1705), he proved in his "Tractatus de seriebus infinitis" that the sum of reciprocals of squares is smaller than 2 :
¡@
¡@

 1 + 1/22 + 1/32 + ... + 1/k2 = 1 + 1/4 + 1/16 + 1/9 + 1/16 + ...  Because 1/4 < 1/3, 1/9 < 1/16, 1/16 < 1/10 and  1/k2 < 1/k(k+1)/2 ,¡@                   1+ 1/4 + 1/9 + 1/16 + ... + 1/k2 + ...                < 1 + 1/3 + 1/6 + 1/10 + 1/15 + ... + 2/k(k+1) + ...  ¡@¡@¡@ [ = 2((1- 1/2) + (1/2-1/3) + (1/3-1/4) + (1/4-1/5)) ...  ¡@¡@¡@   = 2( 1- 1/2 + 1/2 -1/3 + 1/3 - 1/4 + 1/4 - ...)¡@   ¡@¡@¡@ = 2 ]  * The [ ] part is proved by Gottfried Wilhelm Leibniz (1646 - 1716)  ¡@

Although the Bernoulli brothers (Johann and Jakob) managed to confined the sum to less than 2, their attempt to find the exact sum failed. Jakob dispairingly said, " if anybody has discovered the answer that makes us feel so defeated, please contact us, we will be very grateful."
¡@ The answer came in 1734, when mathematical genius Leonhard Euler finally solved the problem.

Leonhard Euler was born in 1707 in Basel. His extraordinary intelligence had been obvious since he was a little child, and he became a student of Johann Bernoulli under his father's arrangement. As a young man full of energy and creativity, it was a tragedy that Euler should lose sight of his right eye in 1730, and later he even turned completely blind. Nevertheless, this did not hinder him in the least from making the great achievement of finding sum of reciprocals of squares, which below is his proof :

¡@
¡@

Sum of 1 + 1/4 + 1/9 + 1/16 +...+ 1/k2 + ...

 ¡@     Combining the knowledge of trigonometry and calculus, the following equation can be derived :¡@ sin x = x - x3/3! + x5/5! - x7/7! + x9/9! - ...  ------------- [1]  Let P(x) be a polynomial with power n, its roots being x= a, x= b, x= c ... and x= d. In other words,  P(a) = P(b) = P(c) = ...= P(d) = 0.  Next, let P(0) = 1, then¡@ P(x) = (1- x/a)(1- x/b)(1- x/c)...(1- x/d)  Substituting x=a into the equation, P(a) = 0 because (1- a/a) = 0.  Similarly, P(b) = P(c) = ...= P(d) = 0 For the condition P(0) = 1, we can simply substitute x=0,  P(0) = (1- 0/a)(1- 0/b)(1- 0/c)...(1- 0/d) = (1)(1)(1)...(1) = 1  So we have proved that P(x) = (1- x/a)(1- x/b)(1- x/c)...(1- x/d) possesses the required properties.  We then go back to the original series 1+1/4+1/9+1/25...+1/k2+...= £k2/6.  Euler first stated the expression f(x) = 1- x2/3! + x4/5! - x6/7! + x8/9! - ...  He noticed that the above expression satifies the condition f(0) = 1. If he could find out the roots for f(x)=0, the expression can be factorized into the formula above.  When x < 0,¡@ f(x) = x(1- x2/3! + x4/5! - x6/7! + x8/9! - ...)/x                               ¡@ = (x - x3/3! + x5/5! - x7/7! + x9/9! - ...)/x                                                 = (sin x)/ x¡@¡@ (from [1])  So as long as x < 0 , solving f(x) is equivalent to solving (sinx)/x, and sinx thus equals 0. We can then see that the roots are x=0, x=+£k , x = +2£k ... Since we know that f(0) = 1, we will remove the root x=0.¡@ Then we will arrive at the expression¡@       f(x)    = (1- x/£k)(1- x/-£k)(1- x/2£k)(1- x/-2£k)(1- x/3£k)(1- x/-3 £k)...    = [(1- x/£k)(1+ x/£k)][(1- x/2£k)(1+ x/2£k)][(1- x/3£k) (1+ x/3£k)]...      This is equivalent to        1 - x2/3! + x4/5! - x6/7! + x8/9! - ...      = [ 1- x2/£k2][ 1- x2/4£k2][ 1- x2/9£k2][ 1- x2/16£k2]...       By multiplying the terms and collecting like terms, we eventually arrive at         1 - x2/3! + x4/5! - x6/7! + x8/9! - ...      = [ 1- x2/£k2][ 1- x2/4£k2][ 1- x2/9£k2][ 1- x2/16£k2]...     = 1 - ( 1/£k2 + 1/4£k2 + 1/9£k2+ 1/16£k2+...)x2 + (...)x4 - ... -1/3! = - ( 1/£k2 + 1/4£k2 + 1/9£k2+ 1/16£k2+... ) 1/6 = 1/£k2 ( 1 + 1/4 + 1/9 + 1/16 + ...) Finally, cross multiplication : *    1 + 1/4 + 1/9 + 1/16 + ... = £k2/6    *

Using a similar method, Euler continued to produce sums of reciprocals of even powers of 4, 6, 8 ... :

£a(4) = £k4/ 90     £a(6) = £k6/ 945    £a(8) = £k8/ 9450    £a(10) =  £k10/ 93555   £a(12) = 691£k12/ 638512875

In general, £a(2n) = k£k2n

Despite this, nothing was known about sums of reciprocals of odd powers up to 1978. Then in a tour de force of elementary though difficult reasoning based on a curious expression for £a(3) in terms of binomial coefficients, R. Apery of France proved that £a(3) is irrational. The proof was controversial, and was referred to as "a mixture of miracles and mysteries". Presumably, £a(5), £a(7), £a(9) ... are also irrational, but it is still unknown.

The remaining question is : does £a(3) = k£k3 ?
And do other sums of reciprocals of odd powers show similar features? Even these questions can be solved, there are bound to be still more and more undiscovered problems waiting to be explored by the rich minds of mathematicians.

¡@
¡@
¡@