Basel problem

This article examines the famous **Basel problem**.

## Introduction

The Basel problem was the problem to evaluate the sum

∑

# k
1

∞

1

k
2

.

{\displaystyle \sum\_{k=1}^\infty\dfrac{1}{k^2}.}

This is also related to the Riemann Zeta function as precisely

ζ
(
2
)
.

{\displaystyle \zeta(2).}

 First, we will prove the sum converges using the integral test: Let

f
(
x
# )

1

x
2

{\displaystyle f(x)=\frac{1}{x^2}}

 for

x
∈
[
1
,
∞
)

{\displaystyle x\in[1,\infty)}

. Clearly

f

{\displaystyle f}

 is positive and decreasing on this interval, so the integral test can be applied.

lim

a
→
∞

∫

1

a

1

x
2

d
x

=

lim

a
→
∞

[

−
1

x

]

1

a

=
1
−

lim

a
→
∞

(

1
a

)

=

{\displaystyle
\begin{align}
 \lim\_{a\to\infty}\int\_{1}^{a} \frac{1}{x^2} \,dx &=\lim\_{a\to\infty}\left[\frac{-1}{x}\right]\_{1}^{a} \\
 &=1 - \lim\_{a\to\infty} \left(\frac{1}{a}\right)\\
 &=1.
\end{align}
}

This is finite, so by the integral test the series converges. The next problem is to what value does it converge?

## Euler's approach

Leonhard Euler was the first one to develop a clever approach to "solve" the problem, though not as rigorously as required. We will first demonstrate this approach and then give a rigorous proof.

The power series for the sine function is

sin
⁡
(
x

# k
0

∞

(
−
1

)
k

x

2
k
1

(
2
k
1
)
!

{\displaystyle \sin(x) = \sum\_{k=0}^\infty(-1)^k\dfrac{x^{2k+1}}{(2k+1)!}}

so it holds that:

sin
⁡
(
x
)

x

=

∑

# k
0

∞

(
−
1

)
k

x

2
k

(
2
k
1
)
!

{\displaystyle \frac{\sin(x)}{x} = \sum\_{k=0}^\infty(-1)^k\dfrac{x^{2k}}{(2k+1)!}}

which we can be written out as

sin
⁡
(
x
)

x

=
1
−

x
2

3
!

x
4

5
!

−

x
6

7
!

⋯

{\displaystyle \frac{\sin(x)}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots}

It is obvious that the coefficient of

x
2

{\displaystyle x^2}

 is

1

/

6

{\displaystyle 1/6}

. We will now obtain this coefficient through a different method. Obviously, the LHS function has zeroes at every integer that is a multiple of

π

{\displaystyle \pi }

, except

0

{\displaystyle 0}

. Through polynomial factoring, we obtain that we can show this as

sin
⁡
(
x
)

x

=
(
x
−
π
)
(
x
π
)
(
x
−
2
π
)
(
x
2
π
)
⋯

{\displaystyle \frac{\sin(x)}{x} = (x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\cdots}

Using the difference of two squares formula, we simplify this to

sin
⁡
(
x
)

x

=
(

x
2

−

π
2

)
(

x
2

−
4

π
2

)
(

x
2

−
9

π
2

)
⋯

{\displaystyle \frac{\sin(x)}{x} = (x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots}

Rearranging factors, we get

sin
⁡
(
x
)

x

=

(

1
−

x
2

π
2

)

(

1
−

x
2

4

π
2

)

(

1
−

x
2

9

π
2

)

⋯

{\displaystyle \frac{\sin(x)}{x} = \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots}

Now, with some simple algebra we obtain that the coefficient of

x
2

{\displaystyle x^2}

 in this expression is given by

∑

# k
1

∞

1

(
k
π

)
2

.

{\displaystyle \sum\_{k=1}^\infty\dfrac{1}{(k\pi)^2}.}

If two polynomials are equal everywhere, then all coefficients must be equal; and this product is equal to the power series expansion, so we deduce that

∑

# k
1

∞

1

(
k
π

)
2

=

1
6

{\displaystyle \sum\_{k=1}^\infty\dfrac{1}{(k\pi)^2}=\frac16}

Multiplying thoroughly by

π
2

{\displaystyle \pi^2}

, we obtain the famous solution to the Basel problem:

∑

# k
1

∞

1

k
2

=

π
2

6

{\displaystyle \sum\_{k=1}^\infty\dfrac{1}{k^2}=\frac{\pi^2}{6}}

However, as we stated, this proof is not rigorous enough, despite finding the correct answer.

## A rigorous proof

Now, since we know the answer, we are ready to give a rigorous proof. Using Fourier analysis, we know that for a function

f

{\displaystyle f}

 that is defined over the interval

[
−
π
,
π
)

{\displaystyle [-\pi,\pi)}

 the Fourier series is given by

a
0

2

∑

# n
1

∞

a
n

cos
⁡
(
n
x
)

∑

# n
1

∞

b
n

sin
⁡
(
n
x
)

{\displaystyle \dfrac{a\_0}{2}+\sum\_{n=1}^\infty a\_n\cos(nx)+\sum\_{n=1}^\infty b\_n\sin(nx)}

where

a
0

=

1
π

∫

−
π

π

f
(
x
)

d
x
,

a
n

=

1
π

∫

−
π

π

f
(
x
)
cos
⁡
(
n
x
)

d
x
,

b
n

=

1
π

∫

−
π

π

f
(
x
)
sin
⁡
(
n
x
)

d
x
.

{\displaystyle
\begin{align}
 a\_0&=\dfrac{1}{\pi}\int\_{-\pi}^\pi f(x)\,dx,\\
 a\_n&=\dfrac{1}{\pi}\int\_{-\pi}^\pi f(x)\cos(nx)\,dx,\\
 b\_n&=\dfrac{1}{\pi}\int\_{-\pi}^\pi f(x)\sin(nx)\,dx.
\end{align}
}

We start out by finding the Fourier series for

x
2

{\displaystyle x^2}

.

a
0

=

1
π

∫

−
π

π

x

2

d
# x

2

π

2

3

{\displaystyle a\_0=\dfrac{1}{\pi}\int\limits\_{-\pi}^\pi x^{2}dx=\frac{2\pi^{2}}{3}}

Since the function is even, the sine coefficients will vanish. That is,

b
n

=
0

{\displaystyle b\_n=0}

 for each

n
∈

N

{\displaystyle n \in \N}

. Hence we just need to evaluate

π
⋅

a
n

=

∫

−
π

π

x

2

cos
⁡
(
n
x
)

d
# x
2

∫

0

π

x

2

cos
⁡
(
n
x
)

d
x

{\displaystyle \pi \cdot a\_n=\int\_{-\pi}^\pi x^{2}\cos(nx)\, dx=2\int\_{0}^\pi x^{2}\cos(nx)\, dx}

.

Applying integration by parts yields

π
⋅

a
n

2

=

(

x
2

n

sin
⁡
(
n
x
)

)

0

π

⏟

=
0

(

2
x

n
2

cos
⁡
(
n
x
)

)

0

π

−

(

2

n
3

sin
⁡
(
n
x
)

)

0

π

⏟

=
0

{\displaystyle \frac{\pi\cdot a\_n}{2}=\underbrace{\left(\frac{x^2}{n}\sin(nx)\right)\_0^{\pi}}\_{=0} + \left(\frac{2x}{n^2}\cos(nx)\right)\_0^{\pi}-\underbrace{\left(\frac{2}{n^3}\sin(nx)\right)\_0^{\pi}}\_{=0}}

The sine term vanishes in the Fourier series because we will use integer multiples of pi in the sine function, and that yields zero, and the cosine function yields

(
−
1

)
n

{\displaystyle (-1)^n}

. Thus, we are left with

a
n

=

4
(
−
1

)
n

n
2

.

{\displaystyle a\_n = \frac{4(-1)^n}{n^2}.}

This means we can write the Fourier series as

π
2

3

∑

# n
1

∞

4
(
−
1

)
n

n
2

cos
⁡
(
n
x
)
.

{\displaystyle \dfrac{\pi^2}{3}+\sum\_{n=1}^\infty \frac{4(-1)^n}{n^2}\cos(nx).}

Now, the Fourier convergence theorem states that if

f

{\displaystyle f}

 is defined on

[
−
c
,
c
)

{\displaystyle [-c,c)}

 and piecewise smooth on

[
−
c
,
c
]

{\displaystyle [-c,c]}

, then for every

x

{\displaystyle x}

 the Fourier series of

f

{\displaystyle f}

 on

[
−
c
,
c
)

{\displaystyle [-c,c)}

 converges to a value

s
(
x
)

{\displaystyle s(x)}

. If the

2
c

{\displaystyle 2c}

-periodic extension of

f

{\displaystyle f}

 is continuous at

x

{\displaystyle x}

 then

s
(
x
# )
f
(
x
)

{\displaystyle s(x)=f(x)}

. If it is discontinuous, then it converges to the average of the left and right hand limits of

x

{\displaystyle x}

.

We can use this here as

f
(
x
# )

x
2

{\displaystyle f(x)=x^2}

 is obviously piecewise smooth on the interval

[
−
π
,
π
]

{\displaystyle [-\pi,\pi]}

. Evaluating at

# x
−
π

{\displaystyle x = -\pi}

 we get

s
(
−
π
# )
f
(
−
π
# )

π
2

=

π
2

3

∑

# n
1

∞

4
(
−
1

)
n

n
2

cos
⁡
(
−
π
n
)

⏟

=
(
−

1
n

)

.

{\displaystyle s(-\pi)=f(-\pi)=\pi^2=\dfrac{\pi^2}{3}+\sum\_{n=1}^\infty \frac{4(-1)^n}{n^2}\underbrace{\cos(-\pi n)}\_{=(-1^n)}.}

If we simplify we finally have

π
2

=

π
2

3

4

∑

# n
1

∞

1

n
2

.

{\displaystyle \pi^2=\frac{\pi^2}{3}+4\sum\_{n=1}^{\infty}\frac{1}{n^2}.}

Rearranging for the sum we have the result

∑

# n
1

∞

1

n
2

=

π
2

6

{\displaystyle \sum\_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}}

and this completes the rigorous proof.