[무한급수] $\sum^{\infty}_{n=1}(-1)^{n+1}\frac{1}{n^2} = \frac{\pi^2}{12}$

Problem.

Show that

$$\sum^{\infty}_{n=1}(-1)^{n+1}\frac{1}{n^2} = \frac{\pi^2}{12}.$$

이 문제를 풀기 위해 다음 식을 이용하자.

 

\begin{align*}\sum^{\infty}_{n=1}\frac{1}{n^2}=\frac{\pi^2}{6}\end{align*}

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다음은 Euler's approach를 통한 증명이다.

유한한 다항식의 성질이 무한 급수에서도 성립한다는 것을 가정하고 접근한 것이라 엄밀한 증명(훗날 엄밀하게 증명됨)은 아니지만 직관적인 관찰로 값을 이끌어 낼수 있다는 점에서 흥미롭다.

Euler's Approach.

The Talyor series expansion of the sine function show that

$$sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.$$

Dividing through by x, we have

$$\frac{sin(x)}{x} = 1 - \frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots.$$

Since the roots of $sin(x)/x=0$ occur at $x=n\pi$ where $n=\pm1,\pm2,\pm3,\cdots$ and

we assume we can express this infinite series as a product of linear factors given by its roots,

\begin{align*}\frac{sin(x)}{x}&=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\left(1-\frac{x}{3\pi}\right)\left(1+\frac{x}{3\pi}\right)\cdots\\&=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots.\end{align*}

If we formally multiply out this product and collect all the $x^2$ terms, we see that the $x^2$ coefficient of $sin(x)/x$ is

\begin{align*}-\left(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+\cdots\right)=-\frac{1}{\pi^2}\sum^{\infty}_{n=1}\frac{1}{n^2}.\end{align*}

But from the original infinite series expansion of $sin(x)/x$, the coefficient of $x^2$ is $-1/(3!)=-1/6$. These two coefficients must be equal; thus,

\begin{align*}-\frac{1}{\pi^2}\sum^{\infty}_{n=1}\frac{1}{n^2}=-\frac{1}{6}.\end{align*}

Multiplying through both sides of this equation by $-\pi^2$ give the sum of the reciprocals of the positive square integers.

\begin{align*}\sum^{\infty}_{n=1}\frac{1}{n^2}=\frac{\pi^2}{6}.\end{align*}

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그리고 

[미적분학] The Rearrangement Theorem for Absolutely Convergent Series

 

에 의해 항의 순서를 바꾸어도 식이 성립함을 기억해두자.($\because\sum^{\infty}_{n=1}(-1)^{n+1}\frac{1}{n^2}$는 absolutely convergent series이다.)

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자 이제, 본격적인 접근을 시작하자.

 

Solve. 

Let

\begin{align*}A&=\sum^{\infty}_{n=1}(-1)^{n+1}\frac{1}{n^2}=1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots,\\B&=\sum^{\infty}_{n=1}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots,\\C&=\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots.\end{align*}

Then those series are absolutely convergent.

Dividing through both sides of the equation of A by $1/4$, we have

$$\frac{1}{4}A=\frac{1}{2^2}A=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{6^2}-\frac{1}{8^2}+\cdots.$$

Now, we can express

\begin{align*}\frac{5}{4}A&=A+\frac{1}{4}A\\&=1-0+\frac{1}{3^2}-\frac{2}{4^2}+\frac{1}{5^2}-0+\frac{1}{7^2}-\frac{2}{8^2}+\cdots\\&=1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots-2\left(\frac{1}{4^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\right)\right)\\&=C-\frac{1}{8}B\end{align*}

At a glance, we can notice that $\frac{1}{2}(A+B)=C$.

Thus,

\begin{align*}\frac{5}{4}A&=C-\frac{1}{8}B\\&=\frac{1}{2}(A+B)-\frac{1}{8}B\\&=\frac{1}{2}A+\frac{3}{8}B\end{align*}

Equivalently,

$$A=\frac{1}{2}B\Rightarrow\sum^{\infty}_{n=1}(-1)^{n+1}\frac{1}{n^2}=\frac{\pi^2}{12},$$

$$C=\frac{3}{4}B\Rightarrow\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}.$$