Arithmetic, geometric and partial sums

A sequence lists terms; a series adds them

A sequence assigns a term uₙ to each allowed integer position n. In this lesson the first position is n = 1; a finite sequence stops at a stated final position. The points (n,uₙ) describe discrete positions, not every real input between them.

A series adds terms. Define the nth partial sum by

Subtracting consecutive partial sums cancels their shared terms:

The empty sum is S₀ = 0. With that definition the difference rule works at n = 1 too, but a formula stated only for positive n cannot automatically be used at zero.

For example, if Sₙ = n² + 2n, then for n ≥ 2,

Here S₁ = 3 also agrees with that term formula. The terms 3,5,7 add to S₃ = 15; neither S₃ nor the formula for Sₙ is the third term itself.

Arithmetic progressions have a fixed difference

For first term a and common difference d, reaching the nth term takes n-1 equal changes:

To sum n terms, write them forward and backward. Each paired column adds to first plus last, so 2Sₙ = n(a + uₙ). Therefore

This works for odd n too: the middle term is counted twice when the two copies of the sum are added. The difference d may be positive, zero or negative.

Suppose u₄ = 11 and u₁₀ = 29. Their positions differ by six, so 6d = 29-11 = 18, giving d = 3. Then a = 11-3d = 2. The tenth term is 29, whereas the sum of the first ten terms is 10(2 + 29)/2 = 155. Multiplying the final term by ten would replace the earlier terms by larger ones.

A block from positions m through n contains n-m + 1 terms. Using partial sums,

For example, positions 4 through 9 contain six terms and sum to S₉-S₃, not S₉-S₄.

Geometric progressions have a fixed multiplier

For first term a and common ratio r, each term is the previous term multiplied by r. For nonzero r,

When r = 0, the first term is still a and all later terms are zero. With nonzero terms, a negative r alternates their signs; do not replace it by its magnitude when reconstructing a sequence.

For the finite sum, write Sₙ = a + ar + … + arⁿ⁻¹ and rSₙ = ar + ar² + … + arⁿ. Subtraction cancels the shared powers:

Thus, for r ≠ 1,

If r = 1, every term is a, so Sₙ = na. The quotient formula cannot be used because its denominator would be zero. If a = 0, the sequence and all its partial sums are zero regardless of the chosen multiplier.

Combine finite sums with the same index range

For a finite common range, summation respects addition and subtraction:

For example,

The first component is an arithmetic sum with first term 2 and difference 3; the second is a geometric sum with first term 1 and ratio 2. At n = 3, the combined terms are 1,3,4, giving total 8, which agrees with the formula. Keep both ranges aligned before combining their sums.

Infinite expressions need an extra check: first form the finite partial sums, then ask whether their limit exists. Do not subtract two divergent totals as if infinity were an ordinary number. For instance,

Each of the separate sums of 1 + 2^(-k) and of 1 grows without bound. Their difference is justified at the finite-partial-sum stage, not by writing an undefined subtraction of infinite totals.

An infinite geometric sum needs a convergent ratio

The sum to infinity means the finite limit of Sₙ as the number of terms grows. For nonzero a, a geometric series converges exactly when |r| < 1. Then rⁿ → 0 and

For 3 + 1.5 + 0.75 + …, r = 1/2 and the sum is 6. For 3-1.5 + 0.75-…, r = -1/2 and the sum is 2. With a negative convergent ratio, partial sums alternate around the limit rather than increasing towards it.

At r = 1, nonzero constant terms give partial sums na, with no finite limit. At r = -1, the partial sums alternate between a and zero. For |r| > 1, their magnitudes do not settle. The all-zero series is the exception when a = 0; the ratio condition above was stated for nonzero a.

The remaining tail after n terms is

Its sign may alternate. To bound its size, use |Rₙ| = |a| |r|ⁿ/|1-r|. For the series 3-1.5 + 0.75-…, |Rₙ| = 2(1/2)ⁿ. Eight terms give error 0.0078125 < 0.01, while seven give 0.015625 > 0.01. A small last term and a small remaining total are different quantities.

Term positions remain integers

For uₙ = 100(0.8)ⁿ⁻¹, solving uₙ = 10 gives

The resulting position is not an integer, so 10 is not exactly a term. For the threshold uₙ < 10, dividing by the negative number log(0.8) reverses the inequality: n-1 > 10.319…. The first allowed position is n = 12. Directly, u₁₁ ≈ 10.737 and u₁₂ ≈ 8.590 confirm the boundary.

Do not round a position without checking whether the condition is strict. A non-strict bound n ≥ K permits equality; a strict bound n > K does not, even when K is already an integer.

Before choosing a formula, decide whether the task asks for a term, a finite block, a cumulative total or a limit. Keep the first index, ratio conditions and original integer positions throughout the calculation. Check a proposed threshold at both the chosen position and its neighbour.