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arithmetic-geometric series
It is well known that a finite geometric series is given by
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(1) |
where in general $q=re^{i\theta}$ is complex. When we are dealing with such sums it is common to consider the expression
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(2) |
which we shall call an arithmetic-geometric series. Let us derive a formula for $H_n(q)$ .
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Subtracting,
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We will proceed to eliminate the right-hand side sums.
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By using (1) and solving for $H_n(q)$ , we obtain
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(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
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that is,
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(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.)
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![$\displaystyle H(q)=\lim_{n\to\infty}H_n(q)=\lim_{n\to\infty}\sum_{k=1}^n kq^k= \lim_{n\to\infty}\bigg[\frac{q}{(1-q)^2}(1-q^n)-\frac{nq^{n+1}}{1-q}\bigg],$](http://images.planetmath.org/cache/objects/8087/js/img7.png)
