Geometric and Weighted Geometric Sums

From Geometric Series to Weighted Sums

A geometric series has terms that change by a constant ratio. Its simple structure makes it useful beyond sums of the form \(1+x+x^2+\cdots\): products of geometric series create coefficients that depend on the index. Those coefficients let us evaluate sums whose terms include factors such as \(n\) or \(n^2\).

The key is to distinguish a finite sum from an infinite series, and to check that a value substituted into a power-series representation lies within its interval of convergence. Once those conditions are met, the representation turns an infinite sum into an algebraic expression.

Definition: A geometric sequence with first term \(a\) and common ratio \(r\) has terms \(a, ar, ar^2,\ldots\). Its finite partial sum has finitely many terms. Its infinite geometric series is the limit of those partial sums, if that limit exists.

Finite and Infinite Geometric Sums

For \(r\ne 1\), the sum of the first \(N+1\) powers, starting at exponent zero, is

1+r+r^2+\cdots+r^N = \sum_{k=0}^{N}r^k = \frac{1-r^{N+1}}{1-r}.

This finite identity follows by multiplying the sum by \(1-r\): all intermediate powers cancel, leaving \(1-r^{N+1}\). It applies to every real \(r\ne1\), whether or not an infinite series would converge. When \(r=1\), the finite sum is \(N+1\).

If \(|r|<1\), then \(r^{N+1}\) tends to zero as \(N\) increases. Taking the limit of the finite formula gives the infinite geometric series:

\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}, \qquad |r|<1.

The condition \(|r|<1\) matters. It ensures that the terms decay to zero and that the partial sums approach the displayed value. At \(r=1\), the partial sums grow without bound; at \(r=-1\), they alternate between two values and do not converge.

Worked Example: A Finite Geometric Sum

Evaluate \(\sum_{k=0}^{5}2^k\). Here \(r=2\), \(N=5\), and \(r\ne1\), so the finite-sum formula applies:

\sum_{k=0}^{5}2^k = \frac{1-2^{6}}{1-2} = \frac{1-64}{-1} = 63.

Check by listing and adding the six terms: \(1+2+4+8+16+32=63\). The result is a finite sum, so the fact that \(|2|>1\) does not prevent using the finite formula.

Worked Example: An Infinite Geometric Sum

Evaluate \(\sum_{n=0}^{\infty}(-1/3)^n\). The common ratio is \(x=-1/3\), and \(|x|=1/3<1\), so the infinite geometric formula is valid:

\sum_{n=0}^{\infty}\left(-\frac13\right)^n = \frac{1}{1-(-1/3)} = \frac{1}{4/3} = \frac34.

As a check, the first four terms sum to \(1-1/3+1/9-1/27=20/27\), and the remaining alternating terms continue to adjust this partial sum toward \(3/4\). Substitution into the geometric formula gives \(1/(1+1/3)=3/4\), as required.

Why Repeated Geometric Factors Produce Weights

To create coefficients that depend on the index, multiply geometric series together. For \(|x|<1\), each factor represents \(1/(1-x)\). In the product of two copies, the coefficient of \(x^n\) counts the ways to write \(n\) as a sum of two nonnegative integers. Those pairs are \((0,n),(1,n-1),\ldots,(n,0)\), so there are \(n+1\).

Consequently, multiplying the two representations gives the repeated-factor identity below. The coefficient count also explains why it is not merely another unweighted geometric series.

Result: For \(|x|<1\),

\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty}(n+1)x^n.

For three factors, the coefficient of \(x^n\) counts triples of nonnegative integers whose sum is \(n\). For each first entry \(j\), the remaining two entries can be chosen in \(n-j+1\) ways. Thus the total count is \(1+2+\cdots+(n+1)=(n+1)(n+2)/2\), giving a second repeated-factor identity:

\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty}\frac{(n+1)(n+2)}{2}x^n, \qquad |x|<1.

These identities are valid for \(|x|<1\), where the geometric representations converge absolutely and their products represent the corresponding functions. In particular, a substitution must be checked against this condition before a sum is evaluated.

Deriving the First Weighted Sum

The two-factor identity has coefficient \(n+1\), while the weight we want is \(n\). Subtract the ordinary geometric series coefficient by coefficient. Since \(n=(n+1)-1\), this gives the formula for the first power of the index.

Theorem: For \(|x|<1\),

\sum_{n=0}^{\infty}n x^n = \frac{x}{(1-x)^2}.

Here is the complete algebraic justification. Both series on the right-hand side of the subtraction converge for \(|x|<1\), so their difference has the difference of their coefficients. The \(n=0\) term in the weighted sum is zero.

\begin{aligned} \sum_{n=0}^{\infty}n x^n &= \sum_{n=0}^{\infty}(n+1)x^n - \sum_{n=0}^{\infty}x^n\\ &= \frac{1}{(1-x)^2}-\frac{1}{1-x}\\ &= \frac{1-(1-x)}{(1-x)^2}\\ &= \frac{x}{(1-x)^2}. \end{aligned}

The condition \(|x|<1\) also guarantees convergence of the weighted series: its term magnitudes are \(n|x|^n\), whose successive-term ratio approaches \(|x|<1\). Thus the substitution produces a genuine sum, not just a formal expression.

Worked Example: A Sum Weighted by the Index

Evaluate \(\sum_{n=1}^{\infty}n/4^n\). Write \(1/4^n=(1/4)^n\), so the substitution is \(x=1/4\). Since \(|1/4|<1\), the weighted formula applies. The \(n=0\) term is zero, so starting at \(n=1\) gives the same value:

\begin{aligned} \sum_{n=1}^{\infty}\frac{n}{4^n} &= \frac{1/4}{(1-1/4)^2}\\ &= \frac{1/4}{(3/4)^2}\\ &= \frac{1/4}{9/16}\\ &= \frac14\cdot\frac{16}{9} = \frac49. \end{aligned}

Check the substituted expression: \(x/(1-x)^2\) at \(x=1/4\) is \((1/4)/(9/16)=4/9\). The first two nonzero terms are \(1/4+2/16=3/8\); the remaining positive terms account for the difference \(4/9-3/8=5/72\).

Deriving the Sum Weighted by the Square of the Index

The three-factor identity contains coefficients \((n+1)(n+2)/2\). A convenient way to extract \(n^2\) is to use \(n^2=n(n-1)+n\). The term \(n(n-1)\) is zero for \(n=0\) and \(n=1\); for \(n\ge2\), its coefficient appears in the three-factor series after multiplying that series by \(x^2\) and multiplying by \(2\).

Indeed, the coefficient of \(x^n\) in \(2x^2/(1-x)^3\) is \(2\cdot n(n-1)/2=n(n-1)\) for \(n\ge2\), and is zero for \(n=0,1\). Add the already established formula for \(\sum n x^n\):

Theorem: For \(|x|<1\),

\sum_{n=0}^{\infty}n^2x^n = \frac{x(1+x)}{(1-x)^3}.

To verify the algebra in full, the repeated-factor representation gives the \(n(n-1)\) part, and the first weighted formula gives the \(n\) part:

\begin{aligned} \sum_{n=0}^{\infty}n^2x^n &= \sum_{n=0}^{\infty}n(n-1)x^n + \sum_{n=0}^{\infty}n x^n\\ &= \frac{2x^2}{(1-x)^3} + \frac{x}{(1-x)^2}\\ &= \frac{2x^2+x(1-x)}{(1-x)^3}\\ &= \frac{x+x^2}{(1-x)^3}\\ &= \frac{x(1+x)}{(1-x)^3}. \end{aligned}

As with the first weighted series, \(|x|<1\) ensures convergence: the ratio of successive magnitudes \(n^2|x|^n\) approaches \(|x|\), which is less than one.

Worked Example: A Sum Weighted by the Square of the Index

Evaluate \(\sum_{n=1}^{\infty}n^2/3^n\). Set \(x=1/3\). Because \(|x|=1/3<1\), the square-weight formula applies. The term at \(n=0\) is zero, so the formula also evaluates the sum beginning at \(n=1\):

\begin{aligned} \sum_{n=1}^{\infty}\frac{n^2}{3^n} &= \frac{(1/3)(1+1/3)}{(1-1/3)^3}\\ &= \frac{(1/3)(4/3)}{(2/3)^3}\\ &= \frac{4/9}{8/27}\\ &= \frac49\cdot\frac{27}{8} = \frac32. \end{aligned}

Check by evaluating the closed form at \(x=1/3\): its numerator is \(4/9\), its denominator is \(8/27\), and their quotient is \(3/2\). The first three terms are \(1/3+4/9+9/27=10/9\), which is less than \(3/2\), as expected for a partial sum of positive terms.

Admissible Substitution and Common Errors

For each power-series representation, substitution means replacing \(x\) by a particular number in both the series and its algebraic expression. It is valid only where the representation converges. For the geometric and repeated-factor series used here, the required condition is \(|x|<1\). A negative value inside this interval is allowed; for example, the first weighted formula at \(x=-1/2\) gives \(\sum_{n=0}^{\infty}n(-1/2)^n=-2/9\).

Common mistake: The coefficient identity for the two-factor representation is \((n+1)x^n\), not \(nx^n\). Treating them as identical would incorrectly claim \(\sum_{n=0}^{\infty}n(1/2)^n=1/(1-1/2)^2=4\). The correct calculation subtracts the unweighted geometric series:

\sum_{n=0}^{\infty}n\left(\frac12\right)^n = \frac{1/2}{(1-1/2)^2} = \frac{1/2}{1/4} = 2.

The discrepancy comes from ignoring the \(+1\) in the repeated-factor coefficient. Another error is substituting \(x=1\) into an infinite-series formula: the denominator may look like an algebraic expression, but the original series does not converge there. A finite geometric identity may still apply at \(x=1\) in its separate form, but that does not make the corresponding infinite series convergent.

Choosing the Right Representation

Before evaluating a sum, identify its coefficient pattern and index range. An unweighted power \(x^n\) points to the geometric representation; a factor \(n\) points to the two-factor representation after subtracting the unweighted series; and a factor \(n^2\) can be assembled from \(n(n-1)+n\). Then check the substitution and simplify the resulting rational expression.

Series pattern	Representation for \(\|x\|<1\)
Unweighted powers	\(\sum_{n=0}^{\infty}x^n=1/(1-x)\)
Index weight	\(\sum_{n=0}^{\infty}n x^n=x/(1-x)^2\)
Squared-index weight	\(\sum_{n=0}^{\infty}n^2x^n=x(1+x)/(1-x)^3\)

These formulas turn coefficient structure into a practical evaluation method. Their reliability depends on two things: matching the coefficient correctly and substituting only an admissible value. With both checks in place, repeated geometric factors provide exact values for a useful family of infinite sums.

Check Your Understanding

For each question, state any convergence condition before evaluating or comparing expressions.

Evaluate the finite sum \(\sum_{k=0}^{4}3^k\). Why does \(|3|>1\) not invalidate the finite-sum formula?
Evaluate \(\sum_{n=0}^{\infty}(-1/4)^n\).
Use a weighted geometric representation to evaluate \(\sum_{n=1}^{\infty}n/5^n\).
Use a repeated-factor representation to evaluate \(\sum_{n=1}^{\infty}n^2/2^n\).
Why can the infinite geometric formula not be used at \(x=1\)?