The Real And Complex Number Systems

1 fp k fp k2 ... ∑fp k n
n0
1
1 − fp k
since |fp k | 1 for all p k . (Suppose NOT, then |fp k | ≥ 1 |fp kn | |fp k | n ≥ 1
contradicts to lim n→ fn 0. ).
Hence,
∑
n1
fn
k1
1
1 − fp k .
8.46 This exercise outlines a simple proof of the formula 2 2 /6. Start with the
inequality sinx x tan x, valid for 0 x /2, taking recipocals, and square each
member to obtain
cot 2 x 1 1 cot
x 2 x.
2
Now put x k/2m 1, wherek and m are integers, with 1 ≤ k ≤ m, and sum on k to
obtain
m
∑ cot 2 k
m
m
2m 12
2m 1
∑
1 m ∑ cot
2 k 2
2
2m k
1 .
k1
k1
k1
Use the formula of Exercise 1.49(c) to deduce the ineqaulity
m2m − 1 2
32m 1 2
m
∑
k1
1
k 2
2mm 12
32m 1 2
Now let m → to obtain 2 2 /6.
Proof: The proof is clear if we follow the hint and Exercise 1.49 (c), soweomitit.
8.47 Use an argument similar to that outlined in Exercise 8.46 to prove that
4 4 /90.
Proof: The proof is clear if we follow the Exercise 8.46 and Exercise 1.49 (c), sowe
omit it.
Remark: (1) From this, it is easy to compute the value of 2s, where
s ∈ n : n ∈ N. In addition, we will learn some new method such as Fourier series and
so on, to find the value of Riemann zeta function.
(2) Ther is an open problem that 2s − 1, wheres ∈ n ∈ N : n 1.