We have got
∞
2
4
1 (2 − 1) 2
( ) =
−
∑
(2 − 1)
=1 Let’s say if = 0 and see what we can derive. First, we know that when = 0 , f(0) = |0| = 0 . Then, we can see that this result can also be expressed in another since we now have the expression in the form of Fourier series.
π 2
4 π
1 1 2
1 3 2
1 5 2
1 7 2
f(0) =
−
(
+
+
+
… )
Obviously, I know you must have found something interesting and beautiful. With surprise and confidence, we write down:
π 2
4 π
1 1 2
1 3 2
1 5 2
1 7 2
0 =
−
(
+
+
+
… )
A beautiful and elegant sum is seen after rearranging the equality: 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 … = π 2 8
I know you must have been amazed by such elegance and beauty but this is not the end. An even more beautiful and better-known result can be derived from the equality above, which is known as the Euler’s sum, calculated by Dr. Euler using
another creative and elegant approach. This problem is what the value is of 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2
+ ⋯
Here is the method.
I let S = 1 1 2
+ 1 2 2
+ 1 3 2
+ 1
+ ⋯
4 2
I rearrange the terms of the sum:
1 1 2
1 3 2
1 5 2
1 7 2
1 2 2
1 4 2
1 6 2
1 8 2
S = (
+
+
+
+ ⋯ ) + (
+
+
+
+ ⋯ )
1 1 2
1 3 2
1 5 2
1 7 2
1 2 2
1 1 2
1 2 2
1 3 2
1 4 2
S = (
+
+
+
+ ⋯ ) +
(
+
+
+
+ ⋯ )
2 8
1 4
S =
+
𝑆
S = 2 6 Thus, at last, the beautiful formula appears: 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 … = 2 6
∞
2 6
1 2
𝑖. .∑
=
1 This example shows the application of Fourier series in number theory and points out a path for us that we can look at number theory through Fourier series.
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