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# arithmetic-geometric series

It is well known that a finite geometric series is given by

$\displaystyle G_{n}(q)=\sum_{{k=1}}^{n}q^{k}=\frac{q}{1-q}(1-q^{n}),\qquad q% \neq 1,$ | (1) |

where in general $q=re^{{i\theta}}$ is complex. When we are dealing with such sums it is common to consider the expression

$\displaystyle H_{n}(q):=\sum_{{k=1}}^{n}kq^{k},\qquad q\neq 1,$ | (2) |

which we shall call an arithmetic-geometric series. Let us derive a formula for $H_{n}(q)$.

$\displaystyle H_{n}(q)=\sum_{{k=1}}^{n}kq^{k},\qquad qH_{n}(q)=\sum_{{k=1}}^{n% }kq^{{k+1}}.$ |

Subtracting,

$\displaystyle(1-q)H_{n}(q)=\sum_{{k=1}}^{n}kq^{k}-\sum_{{k=1}}^{n}kq^{{k+1}}=% \sum_{{k=1}}^{n}kq^{k}-\sum_{{k=2}}^{{n+1}}(k-1)q^{k}=\sum_{{k=1}}^{n}kq^{k}-% \sum_{{k=2}}^{n}(k-1)q^{k}-nq^{{n+1}}.$ |

We will proceed to eliminate the right-hand side sums.

$\displaystyle(1-q)H_{n}(q)=q+\sum_{{k=2}}^{n}q^{k}-nq^{{n+1}}=\sum_{{k=1}}^{n}% q^{k}-nq^{{n+1}}.$ |

By using (1) and solving for $H_{n}(q)$, we obtain

$\displaystyle H_{n}(q)=\sum_{{k=1}}^{n}kq^{k}=\frac{q}{(1-q)^{2}}(1-q^{n})-% \frac{nq^{{n+1}}}{1-q}\>\cdot$ | (3) |

The formula (3) holds in any commutative ring with 1, as long as $(1-q)$ is invertible. If $q$ is a complex number and $|q|<1$, (3) is the partial sum of the convergent series

$\displaystyle H(q)=\lim_{{n\to\infty}}H_{n}(q)=\lim_{{n\to\infty}}\sum_{{k=1}}% ^{n}kq^{k}=\lim_{{n\to\infty}}\bigg[\frac{q}{(1-q)^{2}}(1-q^{n})-\frac{nq^{{n+% 1}}}{1-q}\bigg],$ |

that is,

$\displaystyle H(q)=\sum_{{k=1}}^{\infty}kq^{k}=\frac{q}{(1-q)^{2}},\,\qquad|q|% <1.$ | (4) |

This last result giving the sum of a converging arithmetic-geometric series may be, naturally, obtained also from the sum formula of the converging geometric series, i.e.

$1\!+\!q\!+q^{2}\!+\!q^{3}\!+...\,=\frac{1}{1-q},$ |

when one differentiates both sides with respect to $q$ and then multiplies them by $q$:

$1\!+\!2q\!+\!3q^{2}\!+...\,=\frac{1}{(1\!-\!q)^{2}},$ |

$q\!+\!2q^{2}\!+\!3q^{3}\!+...\,=\frac{q}{(1\!-\!q)^{2}}$ |

(A power series can be differentiated termwise on the open interval of convergence.)

## Mathematics Subject Classification

40C99*no label found*

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