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12.9 Some XVIII-th Century Calculations

The following proofs that

\begin{displaymath}\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}\end{displaymath}


\begin{displaymath}\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}\end{displaymath}

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

\begin{displaymath}\sum_{n=0}^\infty z^n={1\over {1-z}}.\end{displaymath}

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

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



By equating the real and imaginary parts, we get

\begin{displaymath}\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}}.\end{displaymath}

For $r=1$, this yields

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

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

\begin{displaymath}\sum_{n=1}^\infty\cos n\theta=-{1\over 2}.\end{displaymath}


\begin{displaymath}{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).\end{displaymath}

Since two antiderivatives of a function differ by a constant

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

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

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

so $\displaystyle {C={1\over 2}\pi}$ and thus
\sum_{n=1}^\infty{{\sin n\theta}\over n}={1\over 2}(\pi-\theta).
\end{displaymath} (12.79)

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

\begin{displaymath}{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}\end{displaymath}

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

\begin{displaymath}{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).\end{displaymath}

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

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

For $\theta=0$, this says

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

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

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

Subtract the second equation from the first to get

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


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

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

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

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.

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

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

\begin{displaymath}N=1+{1\over 2}+{1\over 3}+{1\over 4}+{1\over 5}+ \&c.\end{displaymath}


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

Subtract the second series from the first to get

1&=&\left(1-{1\over 2}\right)+\left({1\over 2}-{1\over
...{1\over {2\cdot 3}}+{1\over {3\cdot 4}}+{1\over
{4\cdot 5}}+\&c.


\begin{displaymath}1={1\over {1\cdot 2}}+{1\over {2\cdot 3}}+{1\over {3\cdot 4}}+{1\over {4\cdot

12.81   Exercise. $\quad$
a) Explain why Bernoulli's argument is not valid.
b) Give a valid argument proving that

\begin{displaymath}\sum_{n=1}^\infty {1\over {n(n+1)}}=1.\end{displaymath}

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)$\;$.

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Next: Bibliography Up: 12. Power Series Previous: 12.8 Proof of the   Index