Finite sums and geometric limits

A series adds terms of a sequence

A sequence lists u₀, u₁, u₂, ... . Its N-term partial sum adds exactly indices 0 through N−1. An infinite series sum means the limit of those partial sums, if that finite limit exists; it is not a request to carry out infinitely many additions

How many terms are included from u₄ through u₁₁, inclusive?

Arithmetic sums pair endpoints

For uₖ = a + kd, write a finite block forwards and backwards. Corresponding terms always add to the first plus the last. Twice the sum is N times that pair total, so the sum is N times the average of the endpoints. This works for odd N and negative or zero d.

For u₀ = 4 and d = 3, sum the first seven terms

Sum a later block by its endpoints

For indices m through m + N − 1, the first value is a + md and the last is a + (m + N − 1)d. There are N terms, regardless of m. For a = 10, d = −2 and indices 3 through 7, the values are 4,2,0,−2,−4 and their sum is zero

Prove an arithmetic sum, ungraded

Derive the first-N-term arithmetic sum by reversing and adding the list. Explain why odd N causes no difficulty. Then verify the formula by induction on positive integer N. Independent derivation and proof are ungraded

Model: pairing and induction

Adding the forwards and reversed lists gives N pairs, each equal to 2a + (N−1)d, so 2S_N = N[2a + (N−1)d]. For odd N the middle term is simply counted twice in this doubled sum. The formula gives S₁ = a. If it holds for N, add u_N = a + Nd: S_{N+1} = N[2a + (N−1)d]/2 + a + Nd = (N+1)[2a + Nd]/2, which is the required next case

Geometric sums use a shift and subtraction

For ratio r, write the terms in the partial sum, then multiply the sum by r. Subtract the shifted list so all middle terms cancel. If r is not 1, divide by 1−r. If r = 1, add N copies of a directly. N counts terms, not the last index

For u₀ = 12 and r = −1/2, sum the first five terms

For u₀ = −3 and r = 1, how should the first eight terms be summed?

For u₀ = 7 and r = 0, what is the sum of the first six terms?

Alternating terms and alternating sums

With a = 3 and r = −1, the terms alternate between 3 and −3. Partial sums alternate between 3 and 0. Neither sequence has a finite limit, although both are bounded. Being bounded alone does not imply convergence

In the displayed ratio −1 example, what is the sum of the first six terms?

Prove the finite geometric identity, ungraded

Derive (1−r)S_N = a(1−rᴺ) by shifting and subtracting. State why division by 1−r needs r ≠ 1, and explain r = 0 and r = −1 directly from the recurrence. Independent derivation is ungraded

Model: cancellation with the exceptional cases

Write S_N = a + ar + ... + ar^(N−1) and rS_N = ar + ar² + ... + arᴺ. Subtraction leaves a − arᴺ. Divide only if r ≠ 1. For r = 1, S_N = Na. For r = 0, the recurrence gives a followed by zeroes, so S_N = a for N ≥ 1. For r = −1, pairs cancel: S_N = 0 when N is even and a when N is odd. The identity itself is valid at r = 1, but it then says 0 = 0 and cannot determine S_N

Classify term limits by the absolute ratio

First check a = 0: every term is zero for any real ratio. With a ≠ 0, |r| < 1 gives limit zero, including negative ratios and r = 0. Ratio 1 gives the constant limit a. Ratio −1 alternates between a and −a. Ratio greater than 1 diverges in the sign direction of a. Ratio less than −1 alternates with unbounded magnitude and has no single infinite direction

For a = −4 and r = 2, how do the terms behave?

For a = 0 and r = −3, what is the term limit?

Justify the ratio cases, ungraded

Explain the geometric term-limit classification for every real r and starting value a. Separate a = 0 and r = 0, ±1. Justify shrinking powers when 0 < |r| < 1 and growth when |r| > 1, rather than relying on a short plot. Independent proof is ungraded

Model: powers, signs and subsequences

If a = 0, all terms vanish. For 0 < q < 1, write q = 1/(1+c) with c > 0. Bernoulli’s inequality gives 0 < qⁿ ≤ 1/(1+nc), which tends to zero; this proves decay of |arⁿ| when q = |r|. For q > 1, write q = 1+c; then qⁿ ≥ 1+nc grows without bound. Positive r preserves the sign of a, while negative r gives opposite signs on even and odd indices. For r = −1 the two subsequences are distinct constants; for r < −1 they diverge in opposite directions. Handle r = 0 directly by the recurrence and r = 1 as a constant sequence. Bernoulli’s inequality here follows by induction: multiplying 1+nc by 1+c gives 1+(n+1)c+nc² ≥ 1+(n+1)c.

The infinite geometric series uses partial-sum limits

When |r| < 1, the power rᴺ tends to zero, so S_N tends to a/(1−r). This includes alternating decay. For a ≠ 0 and |r| ≥ 1, the terms do not tend to zero, so the series cannot converge. If a = 0, every partial sum is zero for any r. A nonzero constant term sequence converges as a sequence, while its series diverges

For u₀ = 9 and r = −1/2, find the infinite series sum

For a = 5 and r = 1, which statement is correct?

Choose a finite approximation, ungraded

For a = 9 and r = −1/2, derive the absolute error after N terms and find the smallest positive integer N making that error less than 1/20. Explain why the preceding N fails. Independent error-bound reasoning is ungraded

Model: an exact error bound

For a = 9 and r = −1/2, find the smallest positive integer N for which the absolute error is less than 1/20. Subtract S_N = 6[1−(−1/2)ᴺ] from the limiting sum 6. The signed remainder is 6(−1/2)ᴺ, so its magnitude is 6/2ᴺ. The requirement is 2ᴺ > 120. Since 2⁶ = 64 and 2⁷ = 128, the smallest N is 7. The errors at N = 6 and 7 are 3/32 and 3/64; only the latter is below 1/20. This bound controls the infinite tail, not just the last included term