The Infinite Geometric Series Formula

From Finite Sums to an Infinite Sum

A geometric series is built from terms whose successive values are found by multiplying by the same number. For example, in \(6+3+\frac{3}{2}+\frac{3}{4}+\cdots\), each term is half the preceding term. A finite geometric sum is an ordinary finite calculation; an infinite geometric series asks whether the sequence of its partial sums approaches a finite limit.

The distinction matters. A pattern can continue indefinitely without its accumulated sum settling to a number. The infinite geometric series formula gives a sum only when the common ratio is small enough in magnitude for the later terms to have a vanishing total effect.

Here \(a\) and \(r\) are real numbers. The exponent \(n\) begins at zero, so the first term is \(ar^0=a\). The condition on the ratio will be expressed using its magnitude, \(|r|\), because a negative ratio alternates the signs of the terms but can still produce a convergent series.

The Finite Partial-Sum Formula

For \(r\ne 1\), the first \(N\) terms have the finite-sum expression below. The subtraction that produces it is useful because most terms cancel.

Since \(r\ne 1\), division by \(1-r\) is allowed, giving

This is a formula for a finite sum. To obtain an infinite sum, we must take the limit of \(S_N\), and that limit depends on what happens to \(r^N\).

The Infinite Geometric Series Formula

The condition \(|r|<1\) means that repeated powers of \(r\) approach zero: \(r^N\to 0\) as \(N\to\infty\). This is the geometric-sequence limit established in earlier material. Substituting that limit into the finite partial-sum formula proves the theorem.

The denominator is nonzero under this condition: if \(|r|<1\), then \(r\ne 1\). The argument also handles \(a=0\), in which case every term and the sum are zero.

Worked Examples

Why the Formula Fails When \(|r|\ge 1\)

The finite-sum identity does not itself guarantee an infinite sum. Its limiting behavior has to be checked. For a nonzero first term \(a\), every real ratio with \(|r|\ge1\) leads to divergence, but the reasons at the boundary and beyond it are slightly different.

First, if a series converges, its terms must approach zero. To see why, let \(S_N\) be its partial sums. The \(N\)th term is the difference between two consecutive partial sums. If \(S_N\to S\), then \(S_{N-1}\to S\) as well, and therefore \(S_N-S_{N-1}\to S-S=0\). For a geometric series, this difference is \(ar^{N-1}\).

• If \(|r|>1\): when \(a\ne0\), the magnitudes \(|a||r|^{N-1}\) grow rather than approach zero. Thus the terms cannot approach zero, and the partial sums cannot converge.
• If \(r=1\): every term equals \(a\), so \(S_N=Na\). For \(a\ne0\), these partial sums do not approach a finite number. The finite-sum expression with denominator \(1-r\) is unavailable at \(r=1\).
• If \(r=-1\): the terms alternate between \(a\) and \(-a\). For \(a\ne0\), the partial sums alternate between \(a\) and \(0\), so they do not approach one limit.

The exception is \(a=0\). In that case every term is zero and the series converges to zero for every real \(r\). The usual formula \(a/(1-r)\) still cannot be used at \(r=1\), because its denominator is zero; for \(r\ne1\), it evaluates to zero, consistent with the series.

A Common Mistake

A frequent error is to apply \(a/(1-r)\) whenever \(r\ne1\), forgetting the separate convergence condition \(|r|<1\). For example, with \(a=1\) and \(r=2\), that incorrect substitution gives

But the partial sums are \(1\), \(1+2=3\), \(1+2+4=7\), and \(1+2+4+8=15\). The finite formula confirms their behavior: \(S_N=(1-2^N)/(1-2)=2^N-1\), which grows without bound. The algebraic finite-sum formula is valid for every \(N\) when \(r\ne1\); it is the step \(r^N\to0\) that is valid only when \(|r|<1\). A denominator that is nonzero is not, by itself, enough to justify an infinite-series sum.

Why the Condition Matters

The infinite geometric series formula is useful whenever a quantity is repeatedly scaled by a fixed factor and the resulting contributions are added. Its condition is also a reminder of what an infinite sum means: it is not simply a pattern with an ellipsis, but the limit of its finite partial sums. Ratios close to zero make later terms shrink rapidly; a negative ratio can still work if its magnitude is less than one. Ratios of magnitude at least one do not produce a finite sum for a nonzero first term.