One of my favorite mathematical proofs (not original) is to show that the

Infinite Sum of 1/(kk) = 1 + 1/4 + 1/9 + 1/16 + 1/25 + = (pi.pi)/6

This problem, sometimes called the Basel problem, is named from the Swiss city where the famous mathematician Jakob Bernoulli first posed it after proving that the harmonic series, 1 + 1/2 + 1/3 + 1/4 + 1/5 + . diverges or does not converge to a sum.

The easiest proof of the Basel problem may be to write sin x as an infinite product of linear factors and also to find the Maclaurin series for sin x. Comparing the coefficients of the x cubed terms from each expansion gives the result! Unfortunately, this proof is not very inspiring.

The proof below requires a number of mathematical techniques that math majors learn as undergraduates and which are generally fun to use and thus delight the person:

We start with sin t expressed as an infinite product of its zeros

(1) sin t = t.(1 t/pi).(1 + t/pi).(1 t/2pi).(1 + t/2pi). or making a change in the variable, let t=pi.y

(2) sin pi.y = pi.y . (1 y).(1 + y) . (2-y)/2 . (2+y)/2 . or

(3) sin pi.y = pi.y.(1 yy) . (4 yy)/4 . (9 yy)/9 . (16 yy)/16 .

Taking the ln of both sides of the equation to turn a product into a sum gives

(4) ln (sin pi.y) = ln (pi) + ln(y) + ln (1 yy) + ln (4 yy) ln4 + ln (9 yy) ln9 + ln (16 yy) ln16 +

Taking the derivative of both sides of the equation gives

(5) pi.cos pi.y /sin pi.y = 0 + 1/y 2y/(1-yy) 2y/(4-yy) 0 2y/(9-yy) 0 2y/(16-yy) 0 -

Dividing the equation by 2y and rearranging gives

(6) 1/(1-yy) + 1/(4-yy) + 1/(9-yy) + 1/(16-yy) + = 1/2yy (pi/2y)cot pi.y

Making another change in the variable, let x= -y/i (or y = -ix) where I is the square root of -1, we then have xx = - yy and thus the equation becomes

(7) 1/(1+xx) + 1/(4+xx) + 1/(9+xx) + 1/(16+xx) + = -1/2xx + (pi/2ix)cot (-i.pi.x)

Let us now work with the last term of equation (7) and express it in one of its known equivalent forms; that is, cot z = cos z/sin z = I (e^2iz + 1)/ (e^2iz 1) where ^ denotes an exponent

Thus, (pi/2ix)cot (-i.pi.x) = (pi/2ix)I(e^2pi.x + 1)/( e^2pi.x 1) = (pi/2x)(e^2pi.x 1 +2)/ (e^2pi.x 1) = pi/2x + pi/ x(e^2pi.x 1) and the right hand side of (7) becomes

-1/2xx + (pi/2ix)cot(-i.pi.x) = -1/2xx + pi/2x + pi/ x(e^2pi.x 1)

= (pi.x-1)/2xx + pi/ x(e^2pi.x 1)

= (pi.x-1)(e^2pi.x 1)/2xx(e^2pi.x 1) + 2x.pi/2xx(e^2pi.x -1)

= (pi.xe^2pi.x e^2pi.x pi.x + 1 + 2x.pi) / (2xxe^2pi.x 2xx)

We now rewrite (7) with this equivalent expression on the right hand side of the equation

(8) 1/(1+xx) + 1/(4+xx) + 1/(9+xx) + 1/(16+xx) + = (pi.xe^2pi.x e^2pi.x pi.x + 1 + 2x.pi) / (2xxe^2pi.x 2xx)

As x gets closer to 0 without limit, the LHS = 1 + 1/4 + 1/9 + 1/16 + while the RHS approaches 0/0

Using LHopitals Rule and simplifying gives the following expression: (2pi.xe^2pi.x pi.e^2pi.x + pi)/ (4pi.xxe^2pi.x + 4xe^2pi.x 4x) and this expression also approaches 0/0 as x gets close to 0

Using LHopitals Rule a second time and simplifying gives: 4pi.pi.pi.xe^2pi.x / (8pi.pi.xxe^2pi.x + 16pi.xe^2pi.x + 4e^2pi.x 4) and this expression also goes to 0/0

Using LHopitals Rule a third time = (4pi.pi.pi.e^2pi.x + 8pi.pi.pi.pi.xe^2pi.x)/ (16pi.pi.pi.xxe^2pi.x + 16pi.pi.xe^2pi.x + 16pi.e^2pi.x + 32pi.pi.xe^2pi.x + 8pi.e^2pi.x) and now as x goes to 0

= 4pi.pi.pi / (16pi + 8pi)

= pi.pi / 6

Its amazing that all the above collapses to a simple expression involving the number pi, 3.14159 , the ratio of the circumference of a circle to its diameter. Where did that come from given the original infinite series? Well, we started the proof by looking at the sine curve which has a 2pi cycle.

What is the link between sine curves and an infinite series? We know that the distribution of prime numbers is linked to the Riemann zeta function and more recently to eigenvalues of a random Hermitian matrix and the world of modern physics and that is even more amazing. Anyway we have shown that

(9) 1 + 1/4 + 1/9 + 1/16 + = pi.pi / 6 first proved by the mathematician Leonhard Euler in 1735.

This result leads naturally to the Riemann zeta function and the Riemann Hypothesis, which has kept serious mathematicians busy for decades. The Basel problem remains a good introduction or gateway to more advanced mathematics.

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