12.9 Some XVIII-th Century Calculations

The following proofs that

$$\sum_{n=1}^\infty{1\over {n^2}}=1+{1\over {2^2}}+{1\over {3^2}}+{1\over {4^2}}+\&c={{\pi^2}\over 6}$$

and

$$\sum_{n=0}^\infty{1\over {(2n+1)^2}}=1+{1\over {3^2}}+{1\over {5^2}}+{1\over {7^2}}+\&c={{\pi^2}\over 8}$$

use XVIII-th century standards or rigor. You should decide what parts are justified. I denote $f'(\theta)$ by $${ {{df}\over {d\theta}}}$$ below. By the geometric series formula,

$$\sum_{n=0}^\infty z^n={1\over {1-z}}.$$

If $z=re^{i\theta}$ where $r>0$, $\theta\in\mbox{{\bf R}}$, then

$$\sum_{n=0}^\infty r^ne^{in\theta}={1\over {1-re^{i\theta}}}\cdot{{1-re^{-i\theta}}\over {1-re^{-i\theta}}}$$

so

$$\sum_{n=0}^\infty\left(r^n\cos(n\theta)+ir^n\sin(n\theta)\rig... ...2}}={{(1-r\cos\theta)+ir\sin\theta}\over {1+r^2-2r\cos\theta}}.$$

By equating the real and imaginary parts, we get

$$\sum_{n=0}^\infty r^n\cos n\theta={{1-r\cos\theta}\over {1+r^... ...nfty r^n\sin n\theta={{r\sin\theta}\over {1+r^2-2r\cos\theta}}.$$

For $r=1$, this yields

$$\sum_{n=0}^\infty\cos n\theta={{1-\cos\theta}\over {2-2\cos\theta}}={1\over 2}.$$

Thus, $${1+\sum_{n=1}^\infty\cos n\theta={1\over 2}}$$, so

$$\sum_{n=1}^\infty\cos n\theta=-{1\over 2}.$$

Hence,

$${d\over {d\theta}}\left(\sum_{n=1}^\infty{{\sin n\theta}\over n}\right)={d\over {d\theta}}\left(-{1\over 2}\theta\right).$$

Since two antiderivatives of a function differ by a constant

$$\sum_{n=1}^\infty{{\sin n\theta}\over n}=-{1\over 2}\theta+C$$

for some constant $C$. When $\theta=\pi$, we get

$$0=\sum_{n=1}^\infty{{\sin n\pi}\over n}=-{1\over 2}\pi+C$$

so $${C={1\over 2}\pi}$$ and thus

$$\sum_{n=1}^\infty{{\sin n\theta}\over n}={1\over 2}(\pi-\theta).$$ (12.79)

For $${\theta={\pi\over 2}}$$, this gives us

$${1\over 1}-{1\over 3}+{1\over 5}-{1\over 7}+{1\over 9} +\&c ={1\over 2}\left(\pi-{\pi\over 2}\right)={\pi\over 4}$$

(which is the Gregory-Leibniz-Madhava formula). We can rewrite (12.79) as

$${d\over {d\theta}}\sum_{n=1}^\infty-{{\cos n\theta}\over {n^2}}={d\over {d\theta}}\left(-{{(\pi-\theta)^2}\over 4}\right).$$

Again, since two antiderivatives of a function differ by a constant, there is a constant $C_1$ such that

$$\sum_{n=1}^\infty {{\cos n\theta}\over {n^2}}={{(\pi-\theta)^2}\over 4}+C_1.$$

For $\theta=0$, this says

$$\sum_{n=1}^\infty {1\over {n^2}}={{\pi^2}\over 4}+C_1,$$

and for $\theta=\pi$, this says

$$\sum_{n=1}^\infty{{(-1)^n}\over {n^2}}=C_1.$$

Subtract the second equation from the first to get

$$\sum_{n=1}^\infty{{1-(-1)^n}\over {n^2}}={{\pi^2}\over 4};$$

i.e.,

$$2+{2\over {3^2}}+{2\over {5^2}}+{2\over {7^2}}+\&c ={{\pi^2}\over 4},$$

and thus

$$1+{1\over {3^2}}+{1\over {5^2}}+{1\over {7^2}}+\&c={{\pi^2}\over 8}.$$ (12.80)

Let $${S=1+{1\over {2^2}}+{1\over {3^2}}+{1\over {4^2}}+{1\over {5^2}}+\&c.}$$ Subtract (12.80) from this to get

$$\begin{eqnarray*} S-{{\pi^2}\over 8}&=&{1\over {2^2}}+{1\over {4^2}}+{1\over {6^... ...4}\left[1+{1\over {2^2}}+{1\over {3^2}}+\&c \right]={1\over 4}S. \end{eqnarray*}$$

Hence, $${{3\over 4}S={{\pi^2}\over 8}}$$, and then $${S={{\pi^2}\over 6}}$$. $\mid\!\mid\!\mid$

An argument similar to the following was given by Jacob Bernoulli in 1689 [31, p 443]. Let

$$N=1+{1\over 2}+{1\over 3}+{1\over 4}+{1\over 5}+ \&c.$$

Then

$$N-1={1\over 2}+{1\over 3}+{1\over 4}+{1\over 5}+{1\over 6}+\&c.$$

Subtract the second series from the first to get

$$\begin{eqnarray*} 1&=&\left(1-{1\over 2}\right)+\left({1\over 2}-{1\over 3}\righ... ...{1\over {2\cdot 3}}+{1\over {3\cdot 4}}+{1\over {4\cdot 5}}+\&c. \end{eqnarray*}$$

Therefore,

$$1={1\over {1\cdot 2}}+{1\over {2\cdot 3}}+{1\over {3\cdot 4}}+{1\over {4\cdot 5}}+\&c.$$

12.81   Exercise. $\quad$

a) Explain why Bernoulli's argument is not valid.

b) Give a valid argument proving that

$$\sum_{n=1}^\infty {1\over {n(n+1)}}=1.$$

12.82   Note. The notation $\pi$ was introduced by William Jones in 1706 to represent the ratio of the circumference to the diameter of a circle[15, vol2, p9]. Both Maple and Mathematica designate $\pi$ by Pi$\;$.

The notation $e$ was introduced by Euler in 1727 or 1728 to denote the base of natural logarithms[15, vol 2, p 13]. In Mathematica $e$ is denoted by E$\;$. In the current version of Maple there is no special name for $e$; it is denoted by exp(1)$\;$.