We give two proofs for the well known series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$.
1st proof:
Note that:
$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{\sin nx}{n} &=\mathfrak{Im}\left [ \sum_{n=1}^{\infty}\frac{e^{int}}{n} \right ] \\
&= -\mathfrak{Im}\left [ \ln \left ( 1-e^{ix} \right ) \right ]\\
&=-\arg \left ( 1-e^{ix} \right ) \\
&=\frac{\pi -x}{2}
\end{aligned}$$
By the fundamental theorem of calculus we have that:
$$ \int_{0}^{\pi}\sum_{n=1}^{\infty}\frac{\sin nx}{n}\, {\rm d}x=\int_{0}^{\pi}\frac{\pi-x}{2}\,{\rm d}x \Leftrightarrow \left[ -\sum_{n=1}^{\infty}\frac{\cos nx}{n^2} \right]_0^\pi =\frac{\pi^2}{4} \Leftrightarrow$$
$$\Leftrightarrow \sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=\frac{\pi^2}{4} , \quad \quad \quad \quad \quad (1) $$
We note that the series $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$ converges absolutely since
$$ \left | \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} \right |\leq \sum_{n=1}^{\infty}\frac{1}{n^2}=\zeta(2)$$
This means that we can arrange its terms. Hence this series can be written as:
$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} &=\left ( -1-\frac{1}{3^2}-\frac{1}{5^2}-\cdots \right )+\left ( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\cdots \right ) \\
&= \sum_{n=0}^{\infty}\frac{1}{\left ( 2n+2 \right )^2}-\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\\
&=\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{\left ( n+1 \right )^2}-\sum_{n=0}^{\infty}\frac{1}{\left ( 2n+1 \right )^2} \\
&=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}-\frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} \\
&=-\frac{2}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}=-\frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^2}
\end{aligned}$$
Returning to $(1)$ we get that:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{4}\Leftrightarrow \frac{3}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{4}\Leftrightarrow \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
2nd proof:
Using the calculus of residues by invoking the kernel $\pi \cot \pi z$. Consider the function $\displaystyle f(z)=\frac{\pi \cot \pi z}{z^2}$ . It is staightforward to note that $f$ has poles at the integers and that the residue at $0$ is $-\dfrac{\pi^2}{3}$. Let $N$ be a positive integer and let $C_N$ be a counter clock wise square contour with vertices $\left ( N+\dfrac{1}{2} \right )\left ( \pm 1 \pm i \right )$.
By the residue theorem we have that:
$$\oint_{C_N}f(z)\,{\rm d}z =2\pi i \left [ 2\sum_{n=1}^{N}\frac{1}{n^2}-\frac{\pi^2}{3} \right ]$$
Letting $N \rightarrow +\infty$ that contour integral vanishes because if $\pi z = x+iy $ then a straight forward calculation shows that $\displaystyle \left | \cot \pi z \right |^2 =\frac{\cos^2 x+\sinh^2 y}{\sin^2 x+\sinh^2 y}$. Now it follows that if $z$ lies on the vertical edges of $C_N$ then $\displaystyle \left | \cot \pi z \right |^2=\frac{\sinh^2 x}{1+\sinh^2 x}<1$ and if $z$ lies on the horizontal edges of $C_N$ then
$$\left | \cot \pi z \right |\leq \frac{1+\sinh^2 \pi \left ( N+1/2 \right )}{\sinh^2 \pi \left ( N+1/2 \right )}=\coth \pi \left ( N+\frac{1}{2} \right )\leq \coth^2 \frac{\pi}{2}$$
This shows that the contour vanishes and the result $\displaystyle
\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ follows immediately.
1st proof:
Note that:
$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{\sin nx}{n} &=\mathfrak{Im}\left [ \sum_{n=1}^{\infty}\frac{e^{int}}{n} \right ] \\
&= -\mathfrak{Im}\left [ \ln \left ( 1-e^{ix} \right ) \right ]\\
&=-\arg \left ( 1-e^{ix} \right ) \\
&=\frac{\pi -x}{2}
\end{aligned}$$
By the fundamental theorem of calculus we have that:
$$ \int_{0}^{\pi}\sum_{n=1}^{\infty}\frac{\sin nx}{n}\, {\rm d}x=\int_{0}^{\pi}\frac{\pi-x}{2}\,{\rm d}x \Leftrightarrow \left[ -\sum_{n=1}^{\infty}\frac{\cos nx}{n^2} \right]_0^\pi =\frac{\pi^2}{4} \Leftrightarrow$$
$$\Leftrightarrow \sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=\frac{\pi^2}{4} , \quad \quad \quad \quad \quad (1) $$
We note that the series $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$ converges absolutely since
$$ \left | \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} \right |\leq \sum_{n=1}^{\infty}\frac{1}{n^2}=\zeta(2)$$
This means that we can arrange its terms. Hence this series can be written as:
$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} &=\left ( -1-\frac{1}{3^2}-\frac{1}{5^2}-\cdots \right )+\left ( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\cdots \right ) \\
&= \sum_{n=0}^{\infty}\frac{1}{\left ( 2n+2 \right )^2}-\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\\
&=\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{\left ( n+1 \right )^2}-\sum_{n=0}^{\infty}\frac{1}{\left ( 2n+1 \right )^2} \\
&=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}-\frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} \\
&=-\frac{2}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}=-\frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^2}
\end{aligned}$$
Returning to $(1)$ we get that:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{4}\Leftrightarrow \frac{3}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{4}\Leftrightarrow \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
2nd proof:
Using the calculus of residues by invoking the kernel $\pi \cot \pi z$. Consider the function $\displaystyle f(z)=\frac{\pi \cot \pi z}{z^2}$ . It is staightforward to note that $f$ has poles at the integers and that the residue at $0$ is $-\dfrac{\pi^2}{3}$. Let $N$ be a positive integer and let $C_N$ be a counter clock wise square contour with vertices $\left ( N+\dfrac{1}{2} \right )\left ( \pm 1 \pm i \right )$.
By the residue theorem we have that:
$$\oint_{C_N}f(z)\,{\rm d}z =2\pi i \left [ 2\sum_{n=1}^{N}\frac{1}{n^2}-\frac{\pi^2}{3} \right ]$$
Letting $N \rightarrow +\infty$ that contour integral vanishes because if $\pi z = x+iy $ then a straight forward calculation shows that $\displaystyle \left | \cot \pi z \right |^2 =\frac{\cos^2 x+\sinh^2 y}{\sin^2 x+\sinh^2 y}$. Now it follows that if $z$ lies on the vertical edges of $C_N$ then $\displaystyle \left | \cot \pi z \right |^2=\frac{\sinh^2 x}{1+\sinh^2 x}<1$ and if $z$ lies on the horizontal edges of $C_N$ then
$$\left | \cot \pi z \right |\leq \frac{1+\sinh^2 \pi \left ( N+1/2 \right )}{\sinh^2 \pi \left ( N+1/2 \right )}=\coth \pi \left ( N+\frac{1}{2} \right )\leq \coth^2 \frac{\pi}{2}$$
This shows that the contour vanishes and the result $\displaystyle
\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ follows immediately.
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