Geometric series (${\sum }_{n=0}^{\infty }a{r}^n$)

A geometric series adds the terms of a geometric progression:Its ordinary infinite sum is the limit of a sequence of partial sums $S_N={\sum }_{j=0}^{N-1}a{r}^j$, rather than a new operation that adds infinitely many terms at once. For $a,r$ that are real numbers or complex numbers, if $|r|<1$, the series is absolutely convergent andWhen $a\ne 0$, this condition on the modulus of $r$ is also necessary: for $|r|\ge 1$ the terms fail to tend to zero. The all-zero series $a=0$ converges for every $r$. The finite geometric series identity and $r^N\to 0$ give the result with almost no extra algebra. ^[ref]

With $a=1$ and $r=1/2$, the partial sums approach $2$ from below. Changing the common ratio to $-1/2$ makes the partial sums approach $2/3$ from alternating sides: their increments still shrink. At $r=-1$, those increments never shrink, and the partial sums remain $0,1,0,1,\dots$. Substituting $r=-1$ into $1/(1-r)$ does not supply the ordinary sum of that divergent series.

The formula is useful because many apparently different problems share a fixed multiplier. A repeating decimal such as $0.\overline{27}$ isA geometric distribution counting failures before the first success normalizes by the same identity: for $0<q<1$, the probabilities $(1-q)q^n$ on $n=0,1,\dots$ add to one. The comparison test for series uses a bound by a decaying geometric progression to establish absolute convergence without an exact formula for the series being studied. The iterated contraction gives a summable orbit result uses precisely this idea for repeated applications of a map.

The remainder of a geometric series gives more than an eventual limit: it states exactly how much an approximation omits. A common ratio close to one still gives a convergent series, but can require many terms for a useful approximation.

If complete bounce distances obey $D_{n+1}=r{D}_n$ with $0\le r<1$, their sum is $D_1/(1-r)$. The first observed distance alone does not identify $r$, and a fixed energy-retention fraction need not imply the same distance-retention fraction when turning-point spring compression varies.

For $|r|<1$, let $S_N$ contain the first $N$ terms, with $S_0=0$. The remainder after those terms isThis is the exact absolute error. A convenient bound replaces the denominator by $1-|r|$, giving $|R_N|\le |a||r{|}^N/(1-|r|)$ through the triangle inequality. The exact denominator is preferable when the sign or complex phase of $r$ is known.

For $a>0$ and $0<r<1$, the omitted fraction of the entire sum is $R_N/(a/(1-r))=r^N$. Capturing at least $99$ percent therefore means $r^N\le 0.01$. Taking logarithms and applying the ceiling function gives the smallest integer $N$:This gives $7,21,90,$ and $459$ terms for $r=0.5,0.8,0.95,$ and $0.99$, respectively. The difference is a concrete reason to care about the common ratio, even when all four series converge.

An absolute target has a different scale. For $a=1$, $r=0.8$, and required absolute error at most $0.01$, the total sum is $5$ andThe first acceptable term count is $N=28$: the remainder is approximately $0.0096714$, while at $N=27$ it is approximately $0.0120893$. Counting terms and checking the tail avoids an ambiguous instruction to stop when a term merely looks small.

If a map on a normed vector space satisfies $\Vert T(x)\Vert \le q\Vert x\Vert$ for every $x$, with $0\le q<1$, induction gives $\Vert T^n(x)\Vert \le q^n\Vert x\Vert$. The geometric series then proves absolute convergence of its orbit series whenever the ambient normed vector space is complete. The scalar map $T(x)=\sin (qx)$ satisfies the bound because $|\sin u|\le |u|$.

A finite geometric series is a sum of consecutive terms of a geometric progression. With first term $a$ and $N$ terms, including $N=0$ for the empty sum,Unlike an infinite geometric series, this identity does not need $|r|<1$. For example, $3+6+12+24+48=3(2^5-1)=93$. The number of terms is five; the largest exponent is four.

The cancellation behind the formula is especially useful to remember: multiplying $S_N$ by the common ratio shifts all terms one place. In $S_N-r{S}_N$, every interior term cancels, leaving $a-a{r}^N$. Division by $1-r$ gives the identity when $r\ne 1$; at $r=1$ the sum is directly $Na$. The same argument works for complex numbers.

For integers $m\le n$ and $r\ne 1$, the shifted version isThere are $n-m+1$ terms. All powers must be defined; if a negative exponent is present, $r$ must be nonzero. At $r=1$ this sum is simply $n-m+1$. The finite trigonometric sum uses this identity with a complex exponential as the common ratio.

Finite sums of positive integers in a geometric progression already appear in Euclid's Elements, Book IX, Proposition 35. The argument there compares differences of proportional numbers; Joyce's guide rewrites its conclusion as the modern finite-sum formula. This historical result concerns finite sums, without asserting an infinite limit. ^[ref]

Differentiate the finite geometric series ${\sum }_{n=0}^N{r}^n=(1-r^{N+1})/(1-r)$ with respect to $r$, then multiply by $r$. This proves the title identity for $r\ne 1$ without any infinite-series differentiation. At $r=1$, the sum is $N(N+1)/2$. For $|r|<1$, both $N{r}^N$ and $N{r}^{N+1}$ tend to zero, yielding ${\sum }_{n=1}^{\infty }n{r}^n=r/(1-r{)}^2$.

Taking the real and imaginary parts of the finite geometric series ${\sum }_{m=1}^N{e}^{im\theta }$ yieldsThese formulas apply when $\theta \notin 2\pi \mathbb{Z}$. At multiples of $2\pi$, the sums are directly $N$ and $0$. Complex exponentials turn two real trigonometric summations into one finite geometric series.

For an interval $I$ of $L$ consecutive integers and $e(x)=e^{2\pi ix}$,where $\Vert \beta \Vert$ is the distance to the nearest integer. This follows from the finite geometric-series formula and $|1-e(\beta )|\asymp \Vert \beta \Vert$.