Zeta Sums by Steven Zheng

**By: Tao Steven Zheng (郑涛)**

## **Problem**

Here are three infinite series problems I created that involves the Zeta function

ζ
(
x

# n
1

∞

1

n

x

{\displaystyle \zeta (x)=\sum \_{n=1}^{\infty }{\frac {1}{n^{x}}}}

.

**Problem 1:** Show that

ζ
(
x
# )

2

x

2

x

−
1

∑

# n
0

∞

1

(
2
n
1
)

x

{\displaystyle \zeta (x)={\frac {2^{x}}{2^{x}-1}}\sum \_{n=0}^{\infty }{\frac {1}{{(2n+1)}^{x}}}}

.

**Problem 2:** Given that

ζ
(
2
# )

π

2

6

{\displaystyle \zeta (2)={\frac {{\pi }^{2}}{6}}}

, calculate the series

# n
0

∞

1

(
2
n
1
)

2

{\displaystyle A=\sum \_{n=0}^{\infty }{\frac {1}{{(2n+1)}^{2}}}}

 and

# n
0

∞

1

(
2
n
2
)

2

{\displaystyle B=\sum \_{n=0}^{\infty }{\frac {1}{{(2n+2)}^{2}}}}

.

**Problem 3:** Compute the infinite series

∑

# n
1

∞

(

1

n
(
n
1
)

)

2

{\displaystyle \sum \_{n=1}^{\infty }{\left({\frac {1}{n(n+1)}}\right)}^{2}}

.

## **Solution**

**Problem 1**

ζ
(
x

# n
1

∞

1

n

x

=
1

1

2

x

1

3

x

1

4

x

1

5

x

.
.
.

{\displaystyle \zeta (x)=\sum \_{n=1}^{\infty }{\frac {1}{n^{x}}}=1+{\frac {1}{2^{x}}}+{\frac {1}{3^{x}}}+{\frac {1}{4^{x}}}+{\frac {1}{5^{x}}}+...}

ζ
(
x
# )
1

(

1

2

x

1

4

x

1

8

x

1

16

x

.
.
.

)

(

1

3

x

1

6

x

1

12

x

1

24

x

.
.
.

)

(

1

5

x

1

10

x

1

20

x

1

40

x

.
.
.

)

.
.
.

{\displaystyle \zeta (x)=1+\left({\frac {1}{2^{x}}}+{\frac {1}{4^{x}}}+{\frac {1}{8^{x}}}+{\frac {1}{16^{x}}}+...\right)+\left({\frac {1}{3^{x}}}+{\frac {1}{6^{x}}}+{\frac {1}{12^{x}}}+{\frac {1}{24^{x}}}+...\right)+\left({\frac {1}{5^{x}}}+{\frac {1}{10^{x}}}+{\frac {1}{20^{x}}}+{\frac {1}{40^{x}}}+...\right)+...}

ζ
(
x
# )
1

1

2

x

−
1

2

x

2

x

−
1

∑

# n
0

∞

1

(
2
n
1
)

x

{\displaystyle \zeta (x)=1+{\frac {1}{2^{x}-1}}+{\frac {2^{x}}{2^{x}-1}}\sum \_{n=0}^{\infty }{\frac {1}{{(2n+1)}^{x}}}}

ζ
(
x
# )

2

x

2

x

−
1

2

x

2

x

−
1

∑

# n
1

∞

1

(
2
n
1
)

x

{\displaystyle \zeta (x)={\frac {2^{x}}{2^{x}-1}}+{\frac {2^{x}}{2^{x}-1}}\sum \_{n=1}^{\infty }{\frac {1}{{(2n+1)}^{x}}}}

ζ
(
x
# )

2

x

2

x

−
1

∑

# n
0

∞

1

(
2
n
1
)

x

{\displaystyle \zeta (x)={\frac {2^{x}}{2^{x}-1}}\sum \_{n=0}^{\infty }{\frac {1}{{(2n+1)}^{x}}}}

**Problem 2**

 Calculate:

# n
0

∞

1

(
2
n
1
)

2

{\displaystyle A=\sum \_{n=0}^{\infty }{\frac {1}{{(2n+1)}^{2}}}}

From problem 1:

ζ
(
2
# )

2

2

2

2

−
1

∑

# n
0

∞

1

(
2
n
1
)

2

{\displaystyle \zeta (2)={\frac {2^{2}}{2^{2}-1}}\sum \_{n=0}^{\infty }{\frac {1}{{(2n+1)}^{2}}}}

ζ
(
2
# )

4
3

∑

# n
0

∞

1

(
2
n
1
)

2

{\displaystyle \zeta (2)={\frac {4}{3}}\sum \_{n=0}^{\infty }{\frac {1}{{(2n+1)}^{2}}}}

Since

ζ
(
2
# )

π

2

6

{\displaystyle \zeta (2)={\frac {{\pi }^{2}}{6}}}

, then

∑

# n
0

∞

1

(
2
n
1
)

2

=

3
4

×

π

2

6

{\displaystyle \sum \_{n=0}^{\infty }{\frac {1}{{(2n+1)}^{2}}}={\frac {3}{4}}\times {\frac {{\pi }^{2}}{6}}}

∑

# n
0

∞

1

(
2
n
1
)

2

=

π

2

8

{\displaystyle \sum \_{n=0}^{\infty }{\frac {1}{{(2n+1)}^{2}}}={\frac {{\pi }^{2}}{8}}}

 Calculate:

# n
0

∞

1

(
2
n
2
)

2

{\displaystyle B=\sum \_{n=0}^{\infty }{\frac {1}{{(2n+2)}^{2}}}}

Since

A
# B

π

2

6

{\displaystyle A+B={\frac {{\pi }^{2}}{6}}}

, we have

∑

# n
0

∞

1

(
2
n
2
)

2

π

2

8

=

π

2

6

{\displaystyle \sum \_{n=0}^{\infty }{\frac {1}{{(2n+2)}^{2}}}+{\frac {{\pi }^{2}}{8}}={\frac {{\pi }^{2}}{6}}}

Thus,

∑

# n
0

∞

1

(
2
n
2
)

2

=

π

2

24

{\displaystyle \sum \_{n=0}^{\infty }{\frac {1}{{(2n+2)}^{2}}}={\frac {{\pi }^{2}}{24}}}

**Problem 3**

∑

# n
1

∞

(

1

n
(
n
1
)

)

2

=

∑

# n
1

∞

(

1
n

−

1

n
1

)

2

{\displaystyle \sum \_{n=1}^{\infty }{\left({\frac {1}{n(n+1)}}\right)}^{2}=\sum \_{n=1}^{\infty }{\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)}^{2}}

∑

# n
1

∞

(

1
n

−

1

n
1

)

2

=

∑

# n
1

∞

(

1

n

2

−

2

n
(
n
1
)

1

(
n
1
)

2

)

{\displaystyle \sum \_{n=1}^{\infty }{\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)}^{2}=\sum \_{n=1}^{\infty }\left({\frac {1}{n^{2}}}-{\frac {2}{n(n+1)}}+{\frac {1}{{(n+1)}^{2}}}\right)}

∑

# n
1

∞

1

n

2

=
ζ
(
2
# )

π

2

6

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

∑

# n
1

∞

1

(
n
1
)

2

=
ζ
(
2
)
−
# 1

π

2

6

−
1

{\displaystyle \sum \_{n=1}^{\infty }{\frac {1}{{(n+1)}^{2}}}=\zeta (2)-1={\frac {{\pi }^{2}}{6}}-1}

∑

# n
1

∞

1

n
(
n
1
)

=

∑

# n
1

∞

(

1
n

−

1

n
1

)

=
1

{\displaystyle \sum \_{n=1}^{\infty }{\frac {1}{n(n+1)}}=\sum \_{n=1}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)=1}

Therefore,

∑

# n
1

∞

(

1

n
(
n
1
)

)

2

=
2
×
ζ
(
2
)
−
1
−
2
×
1

{\displaystyle \sum \_{n=1}^{\infty }{\left({\frac {1}{n(n+1)}}\right)}^{2}=2\times \zeta (2)-1-2\times 1}

∑

# n
1

∞

(

1

n
(
n
1
)

)

2

=
2
×

π

2

6

−
3

{\displaystyle \sum \_{n=1}^{\infty }{\left({\frac {1}{n(n+1)}}\right)}^{2}=2\times {\frac {{\pi }^{2}}{6}}-3}

∑

# n
1

∞

(

1

n
(
n
1
)

)

2

=

π

2

3

−
3

{\displaystyle \sum \_{n=1}^{\infty }{\left({\frac {1}{n(n+1)}}\right)}^{2}={\frac {{\pi }^{2}}{3}}-3}