Kummer’s acceleration method

There are several methods for acceleration of the convergence of a given series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{a}_n=S.$(1)

One of the simplest is the following one due to Kummer (1837).

We suppose that the terms $a_n$ of (1) are nonzero.  Let

$$\sum _{n=1}^{\mathrm{\infty }}{b}_n=C$$

be a series with nonzero terms and the known sum $C$.  We use the limit

$$\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\varrho \ne \mathrm{\hspace{0.33em}0}$$

and the identity

$S=\varrho C+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\left(1-\varrho {\displaystyle \frac{b_n}{a_n}}\right)a_n.$(2)

Thus the original series (1) has attained a new form (2) the convergence of which is faster because of

$$\underset{n\to \mathrm{\infty }}{lim}\left(1-\varrho \frac{b_n}{a_n}\right)=\mathrm{\hspace{0.33em}0}.$$

Example.  For replacing the series

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=S$$

by a faster converging series we may take

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}=:C,$$

which, for its part, can be expressed as the telescoping series

$$C=\sum _{n=1}^{\mathrm{\infty }}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\mathrm{\hspace{0.33em}1}.$$

Now we have  $\varrho =1$,  and using (2) we obtain

$$S=\mathrm{\hspace{0.33em}1}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2(n+1)}.$$

The convergence of this series may accelerated similarly taking e.g.

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)(n+2)}=:C,$$

where now  $C=\frac{1}{4}$;  then we get

$$S=\frac{5}{4}+2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2(n+1)(n+2)}.$$

The procedure may be repeated $N$ times in all, yielding the result

$$S=\sum _{n=1}^N\frac{1}{n^2}+N!\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2(n+1)(n+2)\mathrm{\cdots }(n+N)}.$$

As for the sum of this series, seeRiemann zeta function at $s=2$ (http://planetmath.org/valueoftheriemannzetafunctionats2).