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

##### 1 Answer

#### Explanation:

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

#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

Note that

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

Now the Maclaurin expansion for

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

Equating the coefficient of

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

Multiplying both sides by

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