What is #sum_(n=1)^oo 1/n^2# ?

Finding the sum of this series is the "Basel Problem", first posed in 1644 by Pietro Mengoli and solved by Leonhard Euler in 1734.

My favourite way of looking at it is to consider the function:

#sin(x)/x#

Note that:

#lim_(x->0) sin(x)/x = 1#

and when #n# is any non-zero integer:

#sin(npi)/(npi) = 0/(npi) = 0#

Consider the function:

#f(x) = prod_(n in ZZ, n != 0) (1-x/(npi))#

#color(white)(f(x)) = prod_(n = 1)^oo (1-x/(npi))(1+x/(npi))#

#color(white)(f(x)) = prod_(n = 1)^oo (1-x^2/(n^2pi^2))#

#color(white)(f(x)) = 1-1/(pi^2)(sum_(n=1)^oo 1/n^2)x^2+O(x^4)#

From its definition, we find:

#{(f(0) = 1), (f(npi) = 0 " for " n in ZZ " with " n != 0) :}#

We can tell that #f(x)# is well defined for any real value of #x# since given any #x# then for all #n > abs(x)/pi#, we have #0 <= (1-x^2/(n^2pi^2)) < 1#. So the product definition converges and so does the Maclaurin expansion.

Note that #f(x)# matches #sin(x)/x# as #x->0# and for #x = npi#.

Hence by the Weierstrass Factorisation Theorem, they are equal functions.

Now the Maclaurin expansion for #sin(x)/x# is:

#1/(1!)-x^2/(3!)+x^4/(5!)-x^6/(7!)+... = 1-x^2/6+O(x^4)#

Equating the coefficient of #x^2# for this expansion and the one we found for #f(x)#, we find:

#1/6 = 1/pi^2 sum_(n=1)^oo 1/n^2#

Multiplying both sides by #pi^2# and transposing, we find:

#sum_(n=1)^oo 1/n^2 = pi^2/6#