∞
X
π2
1
=
Proof that
2
n
6
n=1
The proof follows Proof 11 of Evaluating ζ(2) at
http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf
AB3 page 33 shows that if
π/2
Z
sinn x dx
In =
0
then
I2n =
1 3 5
2n − 1 π
· · · ··· ·
·
2 4 6
2n
2
It follows that
1 · 2 · 3 · · · · · (2n − 1) · 2n
(2 · 4 · 6 · · · · · 2n)2
1 · 2 · 3 · · · · · (2n − 1) · 2n
=
(2n · 1 · 2 · 3 · · · · · n)2
(2n)! π
= n
4 (n!)2 2
I2n =
Integration by parts gives
Z
π/2
I2n = x cos2n x 0 + 2n
π
2
π
·
2
·
. . . . . . . . . . . . . . . . . . . . . (1)
π/2
x sin x cos2n−1 x dx
0
π/2
= n x2 sin x cos2n−1 x 0 − n
Z
π/2
x2 cos2n x − (2n − 1) sin2 x cos2n−2 x dx
0
= n(2n − 1)Jn−1 − 2n2 Jn
Z π/2
where Jn =
x2 cos2n x dx.
Hence
0
(2n)! π
= n(2n − 1)Jn−1 − 2n2 Jn
4n (n!)2 2
Multiply both sides by
4n (n!)2
to get
2n2 (2n)!
π
4n (n!)2
4n (n!)2
=
n(2n
−
1)J
−
2n2 Jn
n−1
4n2
2n2 (2n)!
2n2 (2n)!
2
4n
n!
4n (n!)2
=
2n(2n − 1)Jn−1 −
Jn
4(2n)! n
(2n)!
4n−1 ((n − 1)!)2
4n (n!)2
=
Jn−1 −
Jn
(2n − 2)!
(2n)!
1
Now we use the telescoping series trick AA3 page 9 and get
N
X
4N (N !)2
π
=
J
−
JN
0
4n2
(2N )!
n=1
We will show that
4N (N !)2
JN = 0
N →∞ (2N )!
from which it will follow that
Z π/2
∞
X
π3
π
2
x
dx
=
=
J
=
0
4n2
24
0
n=1
lim
so, using the multiple rule for series and multiplying both side by 4/π we get
∞
X
π2
1
=
n2
6
n=1
as we wanted.
To show that
4N (N !)2
JN = 0
N →∞ (2N )!
we use the Inequalities for integrals method of AB3 Section 3.1.
We also use the fact that x ≤ π2 sin x for 0 < x < π2 (proved later) with the
Inequality Rule to get
lim
Z
π/2
N 2 cos2N x dx
JN =
0
Z
π/2
≤
0
2
2
π
=
4
Z
π2
=
4
Z
π2
=
4
π
sin x
2
cos2N x dx
π/2
sin2 x cos2N x dx
0
π/2
(1 − cos2 x) cos2N x dx
0
Z
π/2
cos
2N
Z
x dx −
0
!
π/2
cos
2N +2
x dx
0
π2
(I2N − I2N +2 )
4 π2
2N + 1
=
I2N −
I2N
4
2N + 2
π2 1
=
I2N
4 2n + 2
=
2
by AB3 Exercise 2.3(b)
So
4N (N !)2
4N (N !)2 π 2 1
JN ≤
I2N
(2N )!
(2N )! 4 2n + 2
4N (N !)2
π2
I2N
=
8(N + 1) (2N )!
π2
π
=
by (1)
8(N + 1) 2
π3
=
16(N + 1)
Thus
π3 1
16 n + 1
By the Squeeze Rule for sequences it follows that
0 < JN ≤
4N (N !)2
JN = 0
N →∞ (2N )!
lim
and we are done.
Finally we must show
x≤
π
π
sin x for 0 < x <
2
2
which we can do using Strategy 4.1 of AB2. Let
f (x) =
From now on assume 0 < x <
f 0 (x) =
x
π
for 0 < x < .
sin x
2
π
2
sin x − x cos x
sin x
≥ 0 as cos x ≤
by AB1 page 7
2
x
x
so f is increasing on (0, π2 ) and hence, on this interval,
f (x) ≤ f
Thus
π
2
=
π
π/2
= .
sin π/2
2
x
π
≤
sin x
2
so
x≤
π
sin x.
2
3