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Mathematics / geometric-series

Geometric Series

A geometric series sums terms related by a constant multiplier; its infinite form converges when the multiplier’s absolute value is less than one.

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A geometric series is a sum whose successive terms are obtained by multiplying the preceding term by a fixed number, called the common ratio. With initial term aa and common ratio rr, it has the form

a+ar+ar2+ar3+⋯ .a+ar+ar^2+ar^3+\cdots.

A finite geometric series contains a specified number of terms; an infinite series is interpreted through the limit of its partial sums. For a≠0a\ne0, the infinite geometric series converges precisely when ∣r∣<1|r|<1, and its sum is a/(1−r)a/(1-r). (openstax.org)

Definition and notation

The terms of a geometric series form a geometric sequence. The sequence lists the terms, whereas the series adds them. For example, 3,6,12,243,6,12,24 is a geometric sequence with common ratio 22, and 3+6+12+243+6+12+24 is its corresponding finite series. (openstax.org)

Using summation notation, the sum of the first NN terms is

SN=∑n=0N−1arn.S_N=\sum_{n=0}^{N-1}ar^n.

Thus the final term is arN−1ar^{N-1}, not arNar^N. The same expression may be indexed from 11:

SN=∑n=1Narn−1.S_N=\sum_{n=1}^{N}ar^{n-1}.

Changing the indexing does not change the series, provided the exponents and limits are adjusted consistently. (openstax.org)

The initial term and ratio may be real numbers or complex numbers. Defining terms through multiplication, rather than through quotients of consecutive terms, also accommodates r=0r=0: the resulting series is a+0+0+⋯a+0+0+\cdots. The term with exponent zero is understood as the initial term aa. (en.wikipedia.org)

Finite sums and their derivation

For r≠1r\ne1,

SN=a1−rN1−r.\boxed{S_N=a\frac{1-r^N}{1-r}}.

The formula follows from a short algebraic calculation. Multiplying the sum by rr gives

rSN=ar+ar2+⋯+arN.rS_N=ar+ar^2+\cdots+ar^N.

Subtracting this from the original expression cancels all intermediate terms:

(1−r)SN=a−arN.(1-r)S_N=a-ar^N.

Division by 1−r1-r yields the formula. If r=1r=1, every term equals aa, so instead SN=NaS_N=Na. (openstax.org)

For example,

3+6+12+24=31−241−2=45.3+6+12+24 =3\frac{1-2^4}{1-2} =45.

The finite-sum formula applies whether the terms grow or shrink; no condition such as ∣r∣<1|r|<1 is needed for a finite sum. (openstax.org)

Infinite sums and convergence

An infinite series has sum SS when its partial sums have the limit

S=lim⁡N→∞SN.S=\lim_{N\to\infty}S_N.

If ∣r∣<1|r|<1, then rN→0r^N\to0, and the finite-sum formula gives

∑n=0∞arn=a1−r.\boxed{\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}}.

For example,

1+12+14+18+⋯=2,1−12+14−18+⋯=23.1+\frac12+\frac14+\frac18+\cdots=2, \qquad 1-\frac12+\frac14-\frac18+\cdots=\frac23.

These equalities describe limits of finite sums, not the completion of an infinite sequence of additions. (openstax.org)

For a≠0a\ne0, the cases outside the convergence condition are:

  • r=1r=1: the partial sums are NaNa and do not approach a finite limit.
  • r=−1r=-1: the partial sums alternate between aa and 00.
  • ∣r∣>1|r|>1: the terms do not tend to zero.
  • Complex rr with ∣r∣=1|r|=1: the terms have constant nonzero magnitude and therefore cannot tend to zero.

A necessary condition for convergence of any series is that its terms tend to zero. This establishes divergence in the latter cases. The exceptional case a=0a=0 gives the zero series for every ratio. (en.wikipedia.org)

Every convergent geometric series is also absolutely convergent, because

∑n=0∞∣arn∣=∣a∣∑n=0∞∣r∣n=∣a∣1−∣r∣.\sum_{n=0}^{\infty}|ar^n| =|a|\sum_{n=0}^{\infty}|r|^n =\frac{|a|}{1-|r|}.

Consequently, geometric series provide standard comparison series in mathematical analysis. Comparisons with geometric decay underlie the ratio test and root test for more general series. (openstax.org)

Remainders and approximation

Subtracting the finite sum from the infinite sum gives the exact remainder after NN terms:

RN=S−SN=arN1−r,∣r∣<1.R_N=S-S_N=\frac{ar^N}{1-r}, \qquad |r|<1.

Its magnitude is

∣RN∣=∣a∣ ∣r∣N∣1−r∣≤∣a∣ ∣r∣N1−∣r∣.|R_N|=\frac{|a|\,|r|^N}{|1-r|} \leq\frac{|a|\,|r|^N}{1-|r|}.

These expressions follow directly from the two sum formulas and quantify the error made by truncating the series. (openstax.org)

For a fixed ratio, increasing NN multiplies the remainder’s magnitude by ∣r∣|r| at each step. Ratios whose magnitude is close to 11 therefore give slower convergence than ratios of smaller magnitude. For real a>0a>0, positive ratios below 11 produce partial sums that increase toward the sum; negative ratios above −1-1 produce partial sums that alternate around it. (openstax.org)

Power series and calculus

Allowing the ratio to vary gives the fundamental power series identity

11−x=∑n=0∞xn,∣x∣<1.\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n, \qquad |x|<1.

Its radius of convergence is 11; neither real endpoint x=1x=1 nor x=−1x=-1 is included. Substitution generates expansions for other functions, such as

11+x2=∑n=0∞(−1)nx2n,∣x∣<1.\frac{1}{1+x^2} =\sum_{n=0}^{\infty}(-1)^n x^{2n}, \qquad |x|<1.

(openstax.org)

Within the interval of convergence, power series may be differentiated and integrated term by term. Taking the derivative of the geometric expansion produces

1(1−x)2=∑n=1∞nxn−1.\frac{1}{(1-x)^2} =\sum_{n=1}^{\infty}n x^{n-1}.

Taking an integral from 00 to xx produces

−log⁡(1−x)=∑n=1∞xnn,∣x∣<1.-\log(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}, \qquad |x|<1.

These derived series are not themselves geometric: their coefficients vary with nn. (openstax.org)

Repeating decimals

Geometric series express repeating decimals as rational numbers. For example,

0.27‾=27100+271002+271003+⋯=27/1001−1/100=311.0.\overline{27} =\frac{27}{100}+\frac{27}{100^2}+\frac{27}{100^3}+\cdots =\frac{27/100}{1-1/100} =\frac{3}{11}.

More generally, a repeating block of kk digits, interpreted as the integer BB and beginning immediately after the decimal point, represents

0.block‾=B10k−1.0.\overline{\text{block}}=\frac{B}{10^k-1}.

This is an application of the infinite-sum formula with common ratio 10−k10^{-k}. (openstax.org)

Historical geometric interpretation

Archimedes used an argument corresponding to a geometric series in his quadrature of a parabolic segment. Starting with an inscribed triangle of area AA, successive collections of additional triangles contributed areas

A4,A16,A64,….\frac A4,\quad \frac A{16},\quad \frac A{64},\ldots.

In modern notation, the total is

A∑n=0∞(14)n=4A3.A\sum_{n=0}^{\infty}\left(\frac14\right)^n=\frac{4A}{3}.

His proof used the method of exhaustion, rather than modern notation for infinite series, to establish the area of the segment. It provides an early geometric instance of summing quantities that decrease by a fixed proportion. (arxiv.org)

References

  1. 2 Infinite Series - Calculus Volume 2openstax.org
  2. 4 Series and Their Notations - Precalculus 2eopenstax.org
  3. 6 Ratio and Root Tests - Calculus Volume 2openstax.org
  4. Ch. 5 Key Equations - Calculus Volume 2openstax.org
  5. 1 Power Series and Functions - Calculus Volume 2openstax.org
  6. Ch. 6 Key Concepts - Calculus Volume 2openstax.org
  7. Geometric seriesen.wikipedia.org
  8. Archimedes' quadrature of the parabola and minimal coversarxiv.org