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Sum to infinity

Understand when a geometric series converges, find its sum to infinity, calculate signed remainders and error bounds, and convert recurring decimals with worked practice.

Before you startFinite geometric series and logarithmic inequalities

01 / What does an infinite sum mean?

Look at the limit of the partial sums.

We cannot finish adding infinitely many terms one by one. Instead, add the first n terms to get Sₙ, then ask whether those partial sums approach one finite value as n grows without bound. If they do, that limit is the sum to infinity, S∞.

For 3 + 3/2 + 3/4 + …, the partial sums 3, 4.5, 5.25,… approach 6. They do not need to reach 6 at a finite stage.

A sequence and its series ask different questions. The constant sequence 3, 3, 3,… has limit 3, but its partial sums 3, 6, 9,… grow without bound.

Follow the partial sumsExplore

Each dot is a partial sum, not an individual term. The first term is 3. The dashed line is the sum to infinity when it exists. The vertical scale adjusts when you change the ratio.

S₆ = 5.90625, approaching 6.

Remaining signed sum: 0.09375.

The positive terms give partial sums that rise towards 6.

Common ratio Number of terms included 6

Watch alternating partial sums approach a limit

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / The convergence condition

For a nonzero first term, require −1 < r < 1.

A geometric series with a ≠ 0 converges exactly when |r| < 1.

For |r| < 1, rⁿ tends to zero. This includes negative ratios: the partial sums can approach their limit from alternating sides.

For r = 1 and a ≠ 0, Sₙ = na has no finite limit. For r = −1, the partial sums alternate between a and 0. For |r| > 1, the terms do not even tend to zero, so the partial sums cannot converge.

The exception is a = 0: all terms and all partial sums are zero for every r. For r = 0, the first term is a and all later terms vanish, so the sum is a.

03 / Find the sum to infinity

Take the limit of the finite-sum formula.

When |r| < 1, the finite formula Sₙ = a(1 − rⁿ)/(1 − r) tends to a/(1 − r), because rⁿ tends to zero.

S∞ = a/(1 − r), for |r| < 1

9 + 18/5 + 36/25 + …Worked example

a = 9, r = 2/5

The ratio has magnitude below 1.

S∞ = 9/(1 − 2/5) = 15

Use the first term and the ratio of this series.

10 − 5/2 + 5/8 − …Worked example

a = 10, r = −¼

A negative ratio is allowed.

S∞ = 10/[1 − (−¼)] = 8

The denominator is 5/4, not 3/4.

Do not use this formula just because it gives a number. For 2 + 4 + 8 + … it would give −2, but r = 2 fails the condition and the series has no finite sum.

04 / Recover a and r

Use the difference between the infinite and finite sums.

Since a/(1 − r) = S∞, the finite formula becomes Sₙ = S∞(1 − rⁿ). Therefore S∞ − Sₙ = S∞rⁿ. Use this to connect the two supplied totals.

S∞ = 24 and S₃ = 21Worked example

S∞ − S₃ = 24r³ = 3

Subtract the finite sum from the infinite sum.

r³ = 1/8 ⇒ r = ½

An odd power fixes the real sign.

a = 24(1 − ½) = 12

Check |r| < 1 and the supplied totals.

S∞ = 10 and S₂ = 15/2Worked example

10r² = 10 − 15/2 = 5/2 ⇒ r² = ¼

An even power may leave two signs.

r = ½ with a = 5; or r = −½ with a = 15

Both satisfy |r| < 1.

5 + 5/2 = 15/2; 15 − 15/2 = 15/2

Both also give the stated first two-term sum.

Retain every candidate that satisfies both the original equations and the convergence condition.

05 / Parameter restrictions

Apply the condition to the actual common ratio.

First term 7 and ratio 3k − 1Worked example

−1 < 3k − 1 < 1

The first term is nonzero, so |r| < 1 is necessary and sufficient.

0 < k < 2/3

Both endpoints are excluded.

S∞ = 7/(2 − 3k), for 0 < k < 2/3

The formula is only valid on this convergence interval.

If the first term also depends on a parameter, check whether it can be zero separately. For first term k − 2 and ratio 2k − 1, the series converges for 0 < k < 1, and also at k = 2 because every term is then zero. The extra isolated value comes from the zero first term.

06 / The sum still to come

The remainder is signed; the error size is its absolute value.

Rₙ = S∞ − Sₙ = arⁿ/(1 − r)
|Rₙ| = |a| |r|ⁿ / |1 − r|

The remaining tail starts at term n + 1, which is arⁿ, and has the same ratio. This gives another way to see the remainder formula.

First term 6 and ratio −½Worked example

S∞ = 6/(3/2) = 4

The infinite sum is fixed.

S₃ = 6 − 3 + 3/2 = 9/2; R₃ = −1/2

The first three terms overshoot the limit, so the remaining signed sum is negative.

S₄ = 15/4; R₄ = 1/4

The next partial sum is below the limit.

For an accuracy requirement such as “within 0.01”, compare |Rₙ| with 0.01. A negative remainder is not automatically a small error.

07 / Choose enough terms for an accuracy

Solve the error inequality and check adjacent integers.

For 8 + 4 + 2 + …, make the error strictly less than 0.01Worked example

S∞ = 16 and |Rₙ| = 16(½)ⁿ

Use the remainder after n terms, with exponent n.

16(½)ⁿ < 0.01 ⇒ 2ⁿ > 1600

Rearrange using positive quantities.

n > ln 1600/ln 2 ≈ 10.644

The least integer candidate is 11.

|R₁₀| = 0.015625; |R₁₁| = 0.0078125

Thus 11 terms are necessary and sufficient.

Using ln |r| directly also works, but dividing by it reverses the inequality when 0 < |r| < 1. Handle r = 0 directly: after the first term the error is zero.

08 / Recurring decimals as geometric series

Identify the repeating block and its place value.

Convert 0.272727… to a fractionWorked example

0.272727… = 27/100 + 27/10000 + …

The two-digit block repeats every two decimal places.

a = 27/100, r = 1/100

The common ratio depends on the block length.

Sum = (27/100)/(1 − 1/100) = 3/11

The geometric series converges.

Convert 0.233333… to a fractionWorked example

0.233333… = 1/5 + 3/100 + 3/1000 + …

Separate the non-repeating part.

Sum = 1/5 + (3/100)/(1 − 1/10) = 7/30

The recurring tail starts at the hundredths place.

A bar or dots above digits must be read carefully: they specify exactly which block repeats. Keep that block’s starting place separate from its length.

09 / A series of squared terms

Squaring changes both the first term and the ratio.

If the original terms are a, ar, ar²,…, their squares are a², a²r², a²r⁴,… . The new first term is a² and its ratio is r².

Sum of squared terms = a²/(1 − r²), when |r| < 1

Original sum 6 and squared-term sum 12Worked example

a/(1 − r) = 6 ⇒ a = 6(1 − r)

The original series converges.

a²/(1 − r²) = 12

Use r² as the new ratio.

36(1 − r)/(1 + r) = 12 ⇒ r = ½

Cancellation is valid because |r| < 1 excludes r = ±1.

a = 3

A series has first term k − 3 and ratio k. Find all real k for which it converges.

Hint

Check nonzero first terms and a zero first term separately.

Worked solution

For k ≠ 3, convergence requires −1 < k < 1. At k = 3, every term is zero, so it also converges. The full set is (−1, 1) together with {3}.

11 / Recap

An infinite sum is a limit, with a condition.

For nonzero a, a geometric series converges exactly when |r| < 1.
Then S∞ = a/(1 − r).
Rₙ = arⁿ/(1 − r) is signed; accuracy uses |Rₙ|.
Check neighbouring integers for the least number of terms.
Use the repeating block’s place value for decimal conversions.
Squared terms form a new geometric series with ratio r².
The all-zero series converges for every ratio.

Review finite geometric series →

Section 1 of 11 · What does an infinite sum mean?