Uploaded by El-Farouk Omar

# Some Very Challenging Calculus Problems-2

```S OME V ERY C HALLENGING C ALCULUS P ROBLEMS
Joseph Breen
Here are two difficult calculus problems, solved using only (sophisticated and clever applications of)
elementary calculus. In particular, there is no complex analysis or use of the residue theorem, Fourier
series, or anything like that. Both problems were the basis for talks given at the UCLA GSO Seminar.
The integral is the concatenation of two integrals from [3]. The infinite series was originally evaluated
by other methods in [2], and the solution presented below is inspired by the solution from [4], together with
other computations found on the internet and my own computational decisions.
Contents
1
A Really Hard Integral
1
2
A Really Hard Infinite Series
6
1
A Really Hard Integral
The integral we will evaluate is
Z
π
2
I=
arccos
0
cos x
1 + 2 cos x
(1)
dx.
Step 1: Rewrite the integrand with trigonometry and then introduce a double integral.
We begin with some trigonometry. Recall the double angle identity cos(2θ) = 2 cos2 θ − 1. This implies
q
1+α
, and thus
2θ = arccos 2 cos2 θ − 1 . Letting α = 2 cos2 θ − 1 yields θ = arccos
2
r
arccos(α) = 2 arccos
1+α
2
!
.
Using this, the integral becomes
π
2
Z
I=
!
1 + 3 cos x
dx.
2 + 4 cos x
{z
}
r
2 arccos
0
|
θ∗
Next, consider a right triangle with angle θ∗ .
q
1
θ∗
q
1+cos x
2+4 cos x
1+3 cos x
2+4 cos x
This picture implies
r
arccos
1 + 3 cos x
2 + 4 cos x
!
r
= arctan
1
1 + cos x
1 + 3 cos x
!
and so
Z
π
2
r
arctan
I=2
0
1 + cos x
1 + 3 cos x
!
dx.
With the goal of using the aforementioned double angle identity again, we make the substitution x = 2y.
Then dx = 2 dy and we have
r
Z π4
1 + cos 2y
I=4
arctan
dy
1 + 3 cos 2y
0
s
!
Z π4
2 cos2 y
arctan
=4
dy
−2 + 6 cos2 y
0
!
Z π4
cos y
dy.
arctan p
=4
0
2 − 3 sin2 y
|
{z
}
b
Recall that
R
1
a2 +x2
dx =
1
a
arctan( xa )
Z
0
1
+ C. Thus,
Z
1
1 1
1
1
dt
=
dt = arctan(bt)
2
2
2
−2
2
1+b t
b 0 b +t
b
1
= arctan(b).
b
1
0
This implies
Z
π
4
I=4
0
Z
π
4
=4
0
Z
=4
0
π
4
Z 1
cos y
1
p
dt dy
cos2 y
2
2 − 3 sin y 0 1 + 2−3 sin2 y t2
p
Z 1
cos y 2 − 3 sin2 y
dt dy
2
2
2
0 2 − 3 sin y + cos yt
p
Z 1
cos y 2 − 3 sin2 y
2 dt dy.
2
2
0 (t + 2) − (t + 3) sin y
q Next, we adjust constants
q in order to simplify the expression in the numerator. In particular, let sin y =
2
2
3 sin w. Then cos y dy =
3 cos w dw, and the integral becomes
r
√
Z π3 Z 1
2 cos w
2
dt
cos w dw
I=4
2 + 2) − (t2 + 3) 2 sin2 w
3
(t
0
0
3
π
√ Z 3Z 1
cos2 w
=8 3
dt dw
2
2
2
0
0 (3t + 6) − (2t + 6) sin w
π
√ Z 3Z 1
cos2 w
=8 3
dt dw
2
2
2
0
0 t + (2t + 6) cos w
π
√ Z 3Z 1
1
=8 3
dt dw.
2
2
2
0
0 t sec w + (2t + 6)
Step 2: Use a trig substitution, partial fractions, then integration by parts.
Let s = tan w. Then ds = sec2 w dw. Since 1 + tan2 w = sec2 w, we have sec2 w = 1 + s2 and dw =
Thus,
√
√ Z 3Z 1
1
1
I=8 3
dt
ds
2 )t2 + (2t2 + 6)
(1
+
s
1
+
s2
0
0
√
√ Z 3Z 1
1
=8 3
dt ds.
2
2
2 2
0 (1 + s )(3t + t s + 6)
0
2
1
1+s2
ds.
Next, we decompose the integrand with partial fractions. We have
√
√ Z 3Z 1
1
t2
1
−
I=8 3
dt ds.
2
1 + s2
3t2 + t2 s2 + 6
0 2t + 6
0
The terms in the parentheses can be integrated with respect to s using the inverse tangent. Thus, we switch
the order of integration.
√
√ Z 1Z 3
1
1
1
−
I=8 3
ds dt
2t2 + 6 1 + s2
3 + t62 + s2
0
0



√
√ Z 1 1
π
3
1
 −q
 dt.
=4 3
arctan  q
2
3
0 t +3
3 + t62
3 + t62
Here we have again used the fact that
similarly with respect to t. This gives
1
a2 +x2
R
1
a
dx =
arctan( xa ) + C. Next, the
√ Z
√ Z 1
4π 3 1 1
1
1
q
I=
dt
−
4
3
2+3
2 + 3)
3
t
(t
0
0
3+

1
t2 +3
term can be integrated
√

3 
dt
arctan  q
3 + t62
√
Z 1
4π 3 1
1
t
t
1
√ arctan √
√
√
=
arctan
dt
−4
2
3
3
3
t2 + 2
t2 + 2
0 (t + 3)
Z 1
t
2π 2
t
√
=
−4
arctan √
dt.
2
2
9
t2 + 2
0 (t + 3) t + 2
Next, we will integrate by parts. Let
t
u = arctan √
t2 + 2
Then
√
du =
1
1+
t2
t2 +2
&middot;
and
t2 + 2 −
dv =
6
t2
(t2
t
√
dt.
+ 3) t2 + 2
(t2
1
√
dt.
+ 1) t2 + 2
2
√t
t2 +2
t2 + 2
dt =
Next, observe that
p
d
t
1
t
√
arctan( t2 + 2) =
&middot;√
=
.
2
2
dx
(1 + t2 + 2
t +2
(t + 3) t2 + 2
√
Thus, v = arctan( t2 + 2). Integrating by parts with this set up yields


√
1
Z 1
p
2 + 2)
2π 2
t
arctan(
t
√
I=
− 4 arctan √
arctan( t2 + 2) −
dt
2
2
9
t2 + 2
0 (t + 1) t + 2
0
!
√
Z 1
2
2π
π π
arctan( t2 + 2)
√
=
−4
&middot; −
dt
2
2
9
6 3
0 (t + 1) t + 2
√
Z 1
arctan( t2 + 2)
√
dt.
=4
2
2
0 (t + 1) t + 2
Step 3: Differentiate under the integral.
We introduce an additional parameter in the integrand:
√
Z 1
arctan(z t2 + 2)
√
I(z) = 4
dt.
(t2 + 1) t2 + 2
0
3
(2)
We seek I = I(1). By the fundamental theorem of calculus,
Z
Z ∞
0
I (z) dz = lim I(z) − I(1)
⇒
I = lim I(z) −
z→∞
1
z→∞
∞
I 0 (z) dz.
1
We compute each term separately.
1
Z
lim I(z) = 4
z→∞
0
√
limz→∞ arctan(z t2 + 2)
√
dt
(t2 + 1) t2 + 2
π
2
1
Z
√
dt
(t2 + 1) t2 + 2
1
t
= 2π arctan √
t2 + 2
0
=4
0
π2
.
3
=
The second to last equality comes from our previous computation in (2).
Next,
!
√
Z ∞
Z ∞Z 1
2 + 2)
t
arctan(z
d
√
dt dz
I 0 (z) dz = 4
(t2 + 1) t2 + 2
1
1
0 dz
Z ∞Z 1
p
1
1
√
=4
&middot;
&middot;
t2 + 2 dt dz
2 2
2
2
1
0 (t + 1) t + 2 1 + z (t + 2)
Z ∞Z 1
1
=4
dt dz.
2
2 2
2
1
0 (t + 1)(1 + z t + 2z )
We decompose the integrand with partial fractions.
Z ∞
Z ∞Z 1
z2
1
1
0
−
I (z) dz = 4
dt dz
2
1 + t2
1 + z 2 t2 + 2z 2
1
1
0 1+z
Z
Z ∞
Z 1
1
1
z2
1
=4
dt −
dt dz
2
2 2
2
1 + z2
1
0 1 + z t + 2z
0 1+t
Z 1
Z ∞
π
1
1
−
=4
dt
dz
1
2
1 + z2 4
0 2 + z2 + t
1


Z ∞
∞
1
1
1
 dz
= π arctan(z) − 4
&middot;q
arctan  q
1 + z2
1
1
2 + z12
2 + z12


Z ∞
2
π
1
1
1
 dz.
=
−4
&middot;q
arctan  q
2
4
1
+
z
1
1
2+
2+ 1
z2
z2
Next, let r = z1 . Then dz = − r12 dr, and the above integral becomes
Z
1
∞
π2
−4
4
Z
π2
=
−4
4
Z
I 0 (z) dz =
1
0
1
0
1
1
1
1
√
√
&middot;
arctan
dr
1 + r12
2 + r2
2 + r2 r2
1
1
√
arctan √
dr.
(r2 + 1) 2 + r2
2 + r2
Recall that
Z
I=4
0
1
√
arctan( t2 + 2)
√
dt.
(t2 + 1) t2 + 2
4
The integral in the above expression is very similar to I, with an inverted argument in the inverse tangent.
Motivated by this, we invoke a handy trig identity. Consider the following right triangle:
θ2
β
θ1
1
This implies that
θ1 + θ2 =
π
2
⇒
arctan(β) + arctan
1
π
= .
β
2
Using this identity,
Z
∞
I 0 (z) dz =
1
π2
−4
4
=
π2
4
=
π2
4
Z
1
π
p
1
√
− arctan( t2 + 2) dr
2
2
2
0 (r + 1) 2 + r
√
Z 1
Z 1
1
arctan( r2 + 2)
√
√
− 2π
dr + 4
dr
2
2
2
2
0 (r + 1) r + 2
0 (r + 1) r + 2
π2
−
+I
3
where in the last equality we again made use of (2). Thus,
Z ∞
I = lim I(z) −
I 0 (z) dz
z→∞
2 12
π
π
π2
−
−
+I .
=
3
4
3
Solving for I in this last equation gives a final answer of
Z
π
2
arccos
0
cos x
1 + 2 cos x
5
dx =
5π 2
.
24
2
A Really Hard Infinite Series
The infinite series we will evaluate is
2
∞
X
1
1
1
S=
1 + + &middot;&middot;&middot; +
.
n2
2
n
n=1
We adopt the shorthand notation Hn = 1 +
Hn ∼ log n, so S clearly converges.
1
2
+ &middot;&middot;&middot; +
1
n
(3)
to denote the nth harmonic number. Recall that
Step 1: Decompose S into two sums.
1
. Thus,
We begin by rewriting the summand. Observe that Hn+1 = Hn + n+1
1
Hn
Hn2 = Hn Hn+1 −
= Hn Hn+1 −
n+1
n+1
1
1
= Hn Hn+1 −
Hn+1 −
n+1
n+1
Hn+1
1
= Hn Hn+1 −
+
.
n + 1 (n + 1)2
The next step is seemingly out of nowhere. We introduce two infinite series with the eventual goal of using
summation by parts. We have
!
!
n
n
n
n
X
X
X
Hn+1
1
Hk X Hk
1
1
2
Hn = Hn Hn+1 −
+
+
−
+
−
n + 1 (n + 1)2
k
k
k2
k2
k=1
k=1
k=1
k=1
!
n+1
n
n
X Hk n+1
X 1
X
Hk X 1
= Hn Hn+1 −
+
+
−
k
k2
k
k2
k=1
k=1
k=1
k=1
!
n
n
n+1
X 1
X
X Hk n+1
Hk X 1
+
+
−
.
= Hn Hn+1 −
k
k2
k
k2
k=2
k=2
k=1
k=1
The last equality comes from the observation that the terms corresponding to k = 1 in the first and second
summations are both 1. Continuing,
!
n+1
n
n
X1
X
1
Hk X 1
2
Hn = Hn Hn+1 −
Hk −
+
−
k
k
k
k2
k=2
k=1
k=1
!
n+1
n
n
X1
X
Hk X 1
= Hn Hn+1 −
&middot; Hk−1 +
−
k
k
k2
k=2
k=1
k=1
!
n+1
n
n
X
X
Hk X 1
= Hn Hn+1 −
(Hk − Hk−1 ) Hk−1 +
−
k
k2
k=2
k=1
k=1
!
n
n+1
n+1
n
X
X
X
Hk X 1
2
= Hn Hn+1 −
Hk Hk−1 +
Hk−1 +
−
k
k2
k=2
k=2
k=1
k=1
!
n+2
n+1
n
n
X
X
X
Hk X 1
2
−
= Hn Hn+1 −
Hk−1 Hk−2 +
Hk−1 +
k
k2
k=3
k=2
k=1
k=1
!
n+2
n+1
n
n
X
X
X
Hk X 1
2
Hk−1 Hk−2 +
Hk−1 +
= Hn Hn+1 −
−
k
k2
k=3
k=2
k=1
k=1
!
n+1
n
n
X
X
Hk X 1
2
=1+
Hk−1
− Hk−1 Hk−2 +
−
.
k
k2
k=3
k=1
6
k=1
The last equality comes from picking off the last term of the first sum and the first term of the second sum.
Simplifying,
!
n
n+1
n
X
X
Hk X 1
2
Hn = 1 +
−
Hk−1 (Hk−1 − Hk−2 ) +
k
k2
k=1
k=3
k=1
!
n+1
n
n
X Hk−1
X
Hk X 1
=1+
+
−
k−1
k
k2
k=3
k=1
k=1
!
n+1
n
n
X Hk−1
X
Hk X 1
=
+
−
k−1
k
k2
k=2
n
X
=2
k=1
All of this shows that
k=1
k=1
n
Hk X 1
.
−
k
k2
k=1
∞
∞
n
n
∞
X
X
Hn2
1 X Hk X 1 X 1
S=
=2
−
.
n2
n2
k
n2
k2
n=1
n=1
n=1
k=1
k=1
|
{z
} |
{z
}
S1
S2
Step 2: Simplify S1 .
Next we consider S1 . Note that
k
k Z
Hk
1X1
1 X 1 j−1
=
=
x
dx
k
k j=1 j
k j=1 0
1
=
k
Z
1
k
Z
=
1
k
X
xj−1 dx
0 j=1
0
1
1 − xk
dx
1−x
where we have used the partial sum formula for a geometric series in the last equality. Continuing,
Hk
1
=
k
k
Z
1
0
1 − xk
dx =
1−x
Z
0
1
1
1−x
Z
1
tk−1 dt dx.
x
Next, we change the order of integration, taking care to change the bounds appropriately.
Hk
=
k
Z
1
tk−1
0
Z
0
t
1
dx dt = −
1−x
Z
1
tk−1 log(1 − t) dt.
0
We continue simplifying S1 in the same manner. In particular,
n
X
Hk
k=1
k
=−
n Z
X
k=1
Z 1
1
tk−1 log(1 − t) dt
0
1 − tn
log(1 − t) dt
1−t
0
Z 1
Z
log(1 − t) 1 n−1
= −n
y
dy dt
1−t
0
t
Z 1
Z y
log(1 − t)
= −n
y n−1
dt dy.
1−t
0
0
=−
7
(4)
The inner integral can be evaluated by a simple substitution to yield
n
X
Hk
k=1
n
2
=
k
1
Z
y n−1 log2 (1 − y) dy.
0
So
Z
∞
n
∞
X
1 X Hk
1 X 1 1 n−1
y
log2 (1 − y) dy
=
2
n
k
2
n
0
n=1
n=1
k=1
!
Z 1
∞
1
1 X yn
=
log2 (1 − y) dy.
2 0 y n=1 n
Recall that log(1 − y) = −
yn
n=1 n
P∞
for −1 ≤ y &lt; 1. Thus,
Z
∞
n
X
1 X Hk
1 1 log3 (1 − y)
=
−
dy.
n2
k
2 0
y
n=1
k=1
Now we make the change of variables z = 1 − y. This yields
1
−
2
Z
0
1
log3 (1 − y)
1
dy = −
y
2
Z
0
1
log3 z
1
dz = −
1−z
2
Z
1
3
log z
0
∞
X
!
z
n
n=0
∞ Z
1X 1 n
dz = −
z log3 z dz.
2 n=0 0
Next, we integrate by parts three times. First, an observation. By L’Hopital’s rule,
lim
z→0
logm z
1
zj
=0
for any m, j &gt; 0. Thus, if we let u = log3 z and dv = z n dz and integrate by parts, the boundary terms
disappear and we have
Z 1
Z 1
3
z n log3 z dz = −
z n log2 z dz.
n+1 0
0
Integrating by parts two more times yields
1
Z
z n log3 z dz =
0
6
(n + 1)2
Z
1
z n log z dz = −
0
6
(n + 1)3
Z
1
z n dz = −
0
6
.
(n + 1)4
Putting all of this together yields
S1 =
∞
n
∞ Z
∞
∞
X
X
X
1 X Hk
1X 1 n
1
1
3
=
−
z
log
z
dz
=
3
=
3
.
2
4
n
k
2 n=0 0
(n + 1)
n4
n=1
n=0
n=1
k=1
Step 3: Simplify S2 .
(2)
Next, we consider S2 . We adopt the shorthand notation Hn
harmonic number of order 2. Then
∞
(2)
X
Hn
S2 =
.
n2
n=1
= 1+
1
22
+ &middot;&middot;&middot; +
1
n2
to denote the nth
We proceed by summation by parts. We have
N
N N
N
X
X
X
X
1 (2)
(2)
(2)
(2)
(2)
(2) (2)
H
=
1
+
H
−
H
H
=
1
+
H
H
−
Hn−1 Hn(2) .
n
n
n
n
n−1
2 n
n
n=1
n=2
n=2
n=2
8
Next, we reindex the first sum and pull off terms from both in order to simplify.
1+
N
X
Hn(2) Hn(2)
n=2
−
N
X
(2)
Hn−1 Hn(2)
=1+
(2) (2)
HN HN
n=2
1
− 1+
4
1
(2)
= − + HN
4
2
1
(2)
= − + HN
4
2
(2)
= HN
2
−
+
N
X
(2)
HN
2
−
N
X
(2)
(2)
Hn−1 Hn−1 −
n=3
N
X
(2)
Hn−1 Hn(2)
n=3
(2)
(2)
Hn−1 Hn−1 − Hn(2)
−
N
X
(2)
Hn−1 &middot;
n=3
N
X
(2)
+
n=3
(2)
Hn−1 &middot;
n=2
The last inequality comes from the observation that
H1
22
1
n2
1
.
n2
= 14 . Continuing,
(2)
(2)
N
N
2 X
X
Hn−1
Hn −
(2)
−
=
H
N
n2
n2
n=2
n=2
1
n2
N
N
(2)
2 X
X
Hn
1
(2)
−
= HN
+
.
2
4
n
n
n=1
n=1
Putting all of this together and sending N → ∞ yields
∞
X
1
2
n
n=1
S2 =
!2
− S2 +
and so
1
S2 =
2
Step 4: Algebraically relate
P∞
1
n=1 n4
to
∞
X
1
n2
n=1
!2
+
∞
X
1
4
n
n=1
∞
1X 1
.
2 n=1 n4
P∞
1
n=1 n2 .
P∞ 1
n=1 n4 .
Typical arguments use Fourier series, complex analyThere are many ways to evaluate
sis, or infinite product expansions. In the spirit of presenting a complete solution which only uses inteP∞
grals, infinite sums, and ordinary power series, we present a pure algebraic manipulation of n=1 n14 to
P∞ 1
n=1 n2 . This technique comes from https://math.stackexchange.com/q/1006510 and the reference therein. Note that
∞
∞ X
1
1X
2
1
2
=
+ 2 2+ 3
.
n4
5 n=1 n &middot; n3
n &middot;n
n &middot;n
n=1
Let
am,n =
2
1
2
+ 2 2+ 3 .
3
mn
m n
m n
Observe1 that
∞
X
an,n =
n=1
X
am,n −
m,n≥1
=
X
m,n≥1
=
X
X
am,n −
n&gt;m≥1
am,n −
X
m,n≥1
X
X
am,m+n −
m,n≥1
(am,n − am,m+n − am+n,n ) .
m,n≥1
1 Note
that there are some divergent double sums here. It’s probably okay though!
9
am,n
m&gt;n≥1
am+n,n
Now we simplify the above double sequence.
1
2
2
+ 2 2+ 3
am,n − am,m+n − am+n,n =
mn3
m n
m n
1
2
1
2
2
2
+ 2
+ 3
+
+
.
−
−
m(m + n)3
m (m + n)2
m (m + n)
(m + n)n3
(m + n)2 n2
(m + n)3 n
{z
}
|
A
Isolating and simplifying A,
2 3
2m n + mn3 (m + n) + 2n3 (m + n)2 + 2m3 n2 + m3 n(m + n) + 2m3 (m + n)2
A=−
m3 n3 (m + n)3
2 2
2m n (m + n) + mn3 (m + n) + 2n3 (m + n)2 + m3 n(m + n) + 2m3 (m + n)2
=−
m3 n3 (m + n)3
2 2
3
3
2m n + mn + 2n (m + n) + m3 n + 2m3 (m + n)
=−
m3 n3 (m + n)2
2
mn(2mn + n + m2 ) + 2n3 (m + n) + 2m3 (m + n)
=−
m3 n3 (m + n)2
2
3
mn(n + m) + 2n (m + n) + 2m3 (m + n)
=−
m3 n3 (m + n)2
mn(n + m) + 2n3 + 2m3
=−
m3 n3 (m + n)
So
2m2 + mn + 2n2
mn(n + m) + 2n3 + 2m3
−
m3 n3
m3 n3 (m + n)
2m2 (n + m) + 2n2 (n + m) − 2n3 − 2m3
m3 n3 (m + n)
2m2 n + 2n2 m
m3 n3 (m + n)
2m + 2n
2
m n2 (m + n)
2
.
m2 n2
am,n − am,m+n − am+n,n =
=
=
=
=
Putting everything together, we have
∞
∞
X
1X
1
1 X
=
an,n =
(am,n − am,m+n − am+n,n )
4
n
5 n=1
5
n=1
m,n≥1
1 X
2
=
2
5
m n2
m,n≥1
∞ X
∞
X
2
1
1
&middot;
5 m=1 n=1 m2 n2
!2
∞
2 X 1
=
.
5 n=1 n2
=
Step 5: Evaluate
P∞
1
n=1 n2 .
10
P∞
Again, there are numerous ways to evaluate n=1 n12 . We present a solution due to [1] which is thematically consistent with everything else done thus far. Note that
∞
∞
∞
∞
∞
X
X
X
X
1X 1
1
1
1
1
=
+
=
+
.
2
2
2
2
n
(2n)
(2n + 1)
4 n=1 n
(2n + 1)2
n=1
n=0
n=0
n=1
Solving for the desired sum in the above equation gives
∞
∞
X
1
4X
1
=
.
2
n
3
(2n
+
1)2
n=1
n=0
We evaluate the latter sum by converting it to a double integral. Note that
Z 1
∞ Z 1
∞
X
X
1
2n
2n
=
x
y
dx
dy
(2n + 1)2
0
0
n=0
n=0
Z 1Z 1X
∞
(x2 y 2 )n dx dy
=
0
Z
1
0 n=0
1
Z
=
0
0
1
dx dy.
1 − x2 y 2
Next, we make the change of variables
x=
Note that
&quot;
∂u x
∂u y
∂v x
∂v y
&quot;
#
=
sin u
cos v
cos u
cos v
sin u sin v
cos2 u
and
sin v
.
cos u
y=
sin u sin v
cos2 v
cos v
cos u
#
= 1 − tan2 u tan2 v.
It follows that dx dy = (1 − tan2 u tan2 v) du dv. Let E be the region in the the square 0 ≤ u, v ≤ π2 which
is the image of 0 ≤ x, y ≤ 1 under this transformation. Note that this region is defined by sin u ≤ cos v
and sin v ≤ cos u. Note that equality occurs in both inequalities if v = π2 − u. This line divides the square
0 ≤ u, v ≤ π2 into two triangles. The inequalities dictate that E is the triangle with vertices (0, 0), (π/2, 0),
and (0, π/2). Therefore,
Z 1Z 1
ZZ
ZZ
1 − tan2 u tan2 v
π2
1 π π
1
dx
dy
=
&middot;
&middot;
=
.
du
dv
=
du
dv
=
2
2
2 2
2 2 2
8
E 1 − tan u tan v
E
0
0 1−x y
Thus,
∞
∞
X
1
4X
1
4 π2
π2
=
= &middot;
=
.
2
2
n
3 n=0 (2n + 1)
3 8
6
n=1
Step 6: Put it all together.
Summarizing steps 1 through 5 gives the final answer:
!2
∞
∞
∞
∞
X
1
1 X 1
1X 1
11 X 1
1
S = 2S1 − S2 = 6
−
−
=
−
4
2
4
4
n
2 n=1 n
2 n=1 n
2 n=1 n
2
n=1
!2
∞
X
1
n2
n=1
!2
!2
∞
∞
11 2 X 1
1 X 1
=
&middot;
−
2 5 n=1 n2
2 n=1 n2
2
17 π 2
=
.
10 6
Thus,
2
∞
X
1
1
1
17π 4
1
+
+
&middot;
&middot;
&middot;
+
=
.
n2
2
n
360
n=1
11
References
[1] Beukers, F., Calabi, E. Kolk, J. Sums of generalized harmonic series and volumes. Nieuw Arch. Wisk., 1993.
[2] Borwein, D., Borwein, J. On an intriguing integral and some series related to ζ(4). Proceedings of the AMS,
1995.
[3] Nahin, P. Inside Interesting Integrals. Springer, 2015.
[4] Valean, C., Furdui, O. Reviving the quadratic series of Au-Yeung. Classical Journal of Analysis, 2015.
12
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