Zeta(s), Zeta(-1), Euler’s transformations, Accelerating Convergence

\displaystyle {\zeta(s)=\frac{1}{1-2^{1-s}} \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} \sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{c} n \\ k \end{array}\right)(k+1)^{-s}}

We will see how to accelerate the convergence of an alternating series to prove the above formula for {\zeta(s).}

Start with

\displaystyle S=\sum_{n=0}^{\infty} (-1)^n a_n = a_0-a_1+a_2-a_3+a_4-a_5\cdots...

and write it as

\displaystyle S= \frac{a_{0}}{2}+\left(\frac{a_{0}}{2}-\frac{a_{1}}{2}\right)-\left(\frac{a_{1}}{2}-\frac{a_{2}}{2}\right)+\left(\frac{a_{2}}{2}-\frac{a_{3}}{2}\right)-\left(\frac{a_{3}}{2}-\frac{a_{4}}{2}\right)+\dots

by distributing half of each term to two adjacent terms.

For example applying this transformation to

\displaystyle \pi=\sum_{n=0}^{\infty} \frac{(-1)^{n} 4}{2 n+1}=4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\frac{4}{9}-\frac{4}{11}+\frac{4}{13}-\frac{4}{15}+\cdots

we get

\displaystyle \pi =2+\left(2-\frac{2}{3}\right)-\left(\frac{2}{3}-\frac{2}{5}\right)+\left(\frac{2}{5}-\frac{2}{7}\right)-\left(\frac{2}{7}-\frac{2}{9}\right)+\left(\frac{2}{9}-\frac{2}{11}\right)-

\displaystyle \pi =2+\sum_{n=0}^{\infty} \frac{(-1)^{n} 4}{(2 n+1)(2 n+3)}

Applying it again, we get

\displaystyle \pi =2+\frac{2}{3}+\sum_{n=0}^{\infty} \frac{(-1)^{n} 8}{(2 n+1)(2 n+3)(2 n+5)}

After few more times we get,

\displaystyle \pi=2+\frac{2}{3}+\frac{4}{15}+\sum_{n=0}^{\infty} \frac{(-1)^{n} 24}{(2 n+1)(2 n+3)(2 n+5)(2 n+7)}

\displaystyle \pi =2+\frac{2}{3}+\frac{4}{15}+\frac{12}{105}+\sum_{n=0}^{\infty} \frac{(-1)^{n} 96}{(2 n+1)(2 n+3)(2 n+5)(2 n+7)(2 n+9)}

\displaystyle \pi=2+\frac{2}{3}+\frac{4}{15}+\frac{12}{105}+\frac{48}{945}+\sum_{n=0}^{\infty} \frac{(-1)^{n} 480}{(2 n+1)(2 n+3)(2 n+5)(2 n+7)(2 n+9)(2 n+11)}

\displaystyle \pi =2+\frac{2}{3}+\frac{4}{15}+\frac{12}{105}+\frac{48}{945}+ \frac{240}{10395}+\sum_{n=0}^{\infty} \frac{(-1)^{n} 2880}{(2 n+1)(2 n+3)(2 n+5)(2 n+7)(2 n+9)(2 n+11)(2n+13)}

In general,

\displaystyle \pi=\sum_{j=0}^{k-1} \frac{2(j !)}{1 \cdot 3 \cdot 5 \cdots(2 j+1)}+\sum_{n=0}^{\infty} \frac{(-1)^{n} 4(k !)}{(2 n+1)(2 n+3) \cdots(2 n+2 k+1)}

For general sequence the Euler tranform maps the sequence {\displaystyle(a_0, a_1, a_2, \cdots..)} to {\displaystyle(a_0, a_1-a_0, a_2-2a_1+a_0, a_3-3a_2+3a_1-a_0, a_4-a_3+6a_2-4a_1+a_0, \cdots..)}

\displaystyle (a_n) \rightarrow \left(s_n = (-1)^{n}\left(\Delta^{n} a\right)_{0}=\sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{l} n \\ k \end{array}\right) a_{k}\right)

\displaystyle S=\sum_{n=0}^{\infty} (-1)^n a_n = a_0-a_1+a_2-a_3+a_4-a_5\cdots...

\displaystyle S=\frac{a_{0}}{2}+\left(\frac{a_{0}}{2}-\frac{a_{1}}{2}\right)-\left(\frac{a_{1}}{2}-\frac{a_{2}}{2}\right)+\left(\frac{a_{2}}{2}-\frac{a_{3}}{2}\right)-\left(\frac{a_{3}}{2}-\frac{a_{4}}{2}\right)+\dots

\displaystyle S=\frac{1}{2} a_{0}+\frac{1}{4}\left(a_{0}-a_{1}\right)+\frac{1}{4}\left[\left(a_{0}-2 a_{1}+a_{2}\right)-\left(a_{1}-2 a_{2}+a_{3}\right)+\left(a_{2}-2 a_{3}+a_{4}\right)-\ldots\right]

We see that by repeating the process shown in the example that

\displaystyle \sum_{n=0}^{\infty} (-1)^n a_n =\sum_{n=0}^{\infty} \frac{s_n}{2^{n+1}}

Riemann Zeta Formula:

We apply this Euler’s transformation for

\displaystyle \zeta(s) =\frac{1}{1-2^{1-s}} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}

to get

\displaystyle \zeta(s)=\frac{1}{1-2^{1-s}} \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} \sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{c} n \\ k \end{array}\right)(k+1)^{-s}

This expression can be seen to be defined for all values {s \neq 1} although the initial expression only valid for {\Re(s)\ge 0.}

Using this formula we can quickly compute the following

\displaystyle \zeta(0)=-\sum_{n=0}^{\infty} \frac{1}{2^{n+1}} \sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{l} n \\ k \end{array}\right) =-\frac{1}{2}

\displaystyle \zeta(-1)=-\frac{1}{3} \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} \sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{l} n \\ k \end{array}\right)(k+1)=-\frac{1}{3}\left(\frac{1}{2^{0+1}}-\frac{1}{2^{1+1}}\right)=-\frac{1}{12}

\displaystyle \zeta\left(\frac{1}{2}\right)\approx-1.46035450880

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