Jugami Calculus

Jugami › Calculus › Sequences and Series

Sequences and Series

Add up infinitely many numbers and still get an answer.

Can you add up infinitely many numbers? Sometimes the total grows for ever. But sometimes, as with 1/2 + 1/4 + 1/8 + …, the running totals close in on a single number, and that number is the sum. Calculators use the same idea to work out sin x and eˣ: they add up the first few terms of an infinite series.

On this page: sequences and their limits, partial sums with a bar chart you can grow, geometric series, why the harmonic series diverges, three quick convergence tests, Taylor polynomials you can raise degree by degree, six worked examples, nine practice problems and an eight-question self-guided quiz.

The graph of y = sin x in blue with its Taylor polynomials of degree 1, 3, 5 and 7 as dashed curves. Each higher degree hugs the sine wave over a wider stretch around x = 0 before peeling away.y = sin x and its Taylor polynomialsdegrees 1, 3, 5, 7
Taylor polynomials of degree 1, 3, 5 and 7 (dashed) for y = sin x (blue). Each extra term keeps the polynomial on the curve for longer.

Section ISequences and their limits

A sequence is an endless list of numbers a₁, a₂, a₃, …, usually given by a formula for the nth term. For an = 1/n the list is 1, 1/2, 1/3, 1/4, …

A sequence converges to a limit L if its terms get as close to L as you like and stay there. We write limn → ∞an = L. If there is no such number, the sequence diverges. The tricks from limits at infinity still work. For example

limn → ∞3n² + 1 over n² + 5 = limn → ∞3 + 1/n² over 1 + 5/n² = 3

Some sequences diverge by growing without bound, like an = 2n; others never settle, like (−1)n, which flips between −1 and 1. A useful fact: rn goes to 0 when |r| < 1, and diverges when |r| > 1.

Section IISeries and partial sums

A series is what you get when you add the terms of a sequence: Σn = 1∞ an = a₁ + a₂ + a₃ + …. You cannot do infinitely many additions, so instead look at the running totals, the partial sums

Sn = a₁ + a₂ + … + an

The partial sums form a new sequence. If it converges to S, the series converges and its sum is S. If not, the series diverges. Keep the two sequences apart: an is what you add each time, Sn is the total so far.

Section IIIPartial-sum explorer

Section IVGeometric series

In a geometric series each term is the one before times a fixed ratio r: a + ar + ar² + ar³ + …. Multiply the partial sum by r and subtract, and almost everything cancels:

Sn − rSn = a − arn,   so   Sn = a(1 − rn) over 1 − r

When |r| < 1, rn goes to 0, so the series converges and

Σn = 0∞ arn = a over 1 − r,   |r| < 1

When |r| ≥ 1 (and a ≠ 0) it diverges. For 1/2 + 1/4 + 1/8 + …, a = 1/2 and r = 1/2, so the sum is (1/2)/(1/2) = 1: walk half way to a wall, then half the rest, and so on, and the total distance is exactly the distance to the wall.

Section VThe harmonic series diverges

The harmonic series is 1 + 1/2 + 1/3 + 1/4 + …. Its terms shrink to 0, so it is tempting to think the sum settles down. It does not. Group the terms in blocks of 1, 2, 4, 8, … terms:

1 + 1 over 2 + (1 over 3 + 1 over 4) + (1 over 5 + … + 1 over 8) + … ≥   1 + 1 over 2 + 1 over 2 + 1 over 2 + …

Each bracket is at least 1/2, because every term in it is at least as big as its last term: 1/3 + 1/4 ≥ 1/4 + 1/4, and the four terms 1/5 to 1/8 add to at least 4 × 1/8. So S2k ≥ 1 + k/2, and the partial sums pass every number. The growth is very slow: it takes 12,367 terms to pass 10.

The lesson: terms going to 0 does not guarantee that a series converges.

Section VIConvergence tests, in brief

Usually you cannot find a formula for Sn. Tests tell you whether a series converges without finding its sum. Three of the most useful:

  1. The nth-term test. If an does not go to 0, the series diverges. (If it does go to 0, this test says nothing: think of the harmonic series.)
  2. The comparison test. For positive terms with an ≤ bn: if Σ bn converges, so does Σ an; if Σ an diverges, so does Σ bn. Good series to compare with are geometric series and the p-series Σ 1/np, which converges exactly when p > 1.
  3. The ratio test. For positive terms, find L = limn → ∞an+1/an. If L < 1 the series converges; if L > 1 it diverges; if L = 1 the test cannot decide. It works well with powers and factorials.

Section VIIPower series and Taylor polynomials

A power series is a series with a variable in it: Σn = 0∞ cnxn = c₀ + c₁x + c₂x² + …. It is like a polynomial that never ends. Each power series converges on an interval centred at 0 (its radius of convergence can be 0, a positive number, or infinite) and diverges outside it.

Which polynomial best copies a function f near 0? Match the value and as many derivatives at 0 as you can. That gives the Taylor polynomial of degree n about 0 (also called a Maclaurin polynomial):

Pn(x) = f(0) + f′(0)x + f″(0) over 2!x² + … + f(n)(0) over n!xn

Letting n → ∞ gives the Taylor series. Four to know:

ex = 1 + x + x² over 2! + x³ over 3! + …   (all x) sin x = x − x³ over 3! + x⁵ over 5! − …   (all x) cos x = 1 − x² over 2! + x⁴ over 4! − …   (all x) ln(1 + x) = x − x² over 2 + x³ over 3 − …   (−1 < x ≤ 1)

Near 0, a few terms give an excellent approximation. Far from 0, you need more terms, and for ln(1 + x) beyond x = 1 no number of terms will do.

Section VIIITaylor polynomial explorer

Section IXWorked examples

Example 1: the limit of a sequence

Find limn → ∞(5n − 2)/(2n + 7).

(5 − 2/n)/(2 + 7/n)Divide top and bottom by n→ 5/2Answer

✓ a₁₀₀₀ = 4998/2007 ≈ 2.490284, close to 2.5.

Example 2: a geometric sum

Find 6 + 2 + 2/3 + 2/9 + ….

a = 6, r = 1/3Each term is a third of the one before6/(1 − 1/3) = 9Answer

✓ S₅ = 6 + 2 + 0.666667 + 0.222222 + 0.074074 ≈ 8.962963.

Example 3: a repeating decimal

Write 0.272727… as a fraction.

0.27 + 0.0027 + 0.000027 + …A geometric seriesa = 27/100, r = 1/100First term and ratio(27/100)/(99/100) = 27/99 = 3/11Answer

✓ 3 ÷ 11 = 0.272727…

Example 4: the ratio test

Does Σn = 1∞ n/2n converge?

an+1/an = (n + 1)/(2n)Ratio of consecutive terms→ 1/2 < 1Converges by the ratio test

✓ Its sum is 2; the first twenty terms add to about 1.999979.

Example 5: the comparison test

Does Σn = 1∞ 1/(n² + 1) converge?

1/(n² + 1) < 1/n²Smaller positive termsΣ 1/n² converges (p-series, p = 2)CompareSo Σ 1/(n² + 1) convergesComparison test

✓ Its sum is about 1.076674, less than π²/6 ≈ 1.644934.

Example 6: estimating with a Taylor polynomial

Use the degree-3 Taylor polynomial of ex to estimate e0.5.

P₃(x) = 1 + x + x²/2 + x³/6Degree 31 + 0.5 + 0.125 + 0.0208333Substitute x = 0.5≈ 1.645833Estimate

✓ e0.5 ≈ 1.648721, so the error is about 0.002888.

Section XCommon mistakes

  • Mixing up the sequence and the series. 1/n → 0, but Σ 1/n diverges.
  • Using the nth-term test backwards. Terms going to 0 never proves convergence.
  • Using a/(1 − r) when |r| ≥ 1. 1 + 2 + 4 + … does not equal −1.
  • Starting at the wrong term. In a/(1 − r), a is the first term actually in the sum.
  • Trusting a Taylor polynomial far from 0. The error grows as x moves away, and outside the interval of convergence more terms make it worse.

Section XIPractice problems

Try each one on paper first, then open the answer to check your steps.

  1. Find the limit of an = (2n + 1)/(n + 3). Sequence

    Show answer and steps
    (2 + 1/n)/(1 + 3/n)Divide top and bottom by n→ (2 + 0)/(1 + 0) = 2Let n → ∞

    ✓ a₁₀₀₀ = 2001/1003 ≈ 1.995015, already close to 2.

  2. For Σk = 1∞ 1/(k(k + 1)), find S₄ and the sum of the series. Partial sums

    Show answer and steps
    1/(k(k + 1)) = 1/k − 1/(k + 1)Split the fractionS₄ = 1 − 1/5 = 4/5The middle terms cancelSₙ = 1 − 1/(n + 1) → 1Sum of the series: 1

    ✓ 1/2 + 1/6 + 1/12 + 1/20 = 0.8.

  3. Find the sum of 5 − 5/2 + 5/4 − 5/8 + …. Geometric

    Show answer and steps
    a = 5, r = −1/2First term and ratio5/(1 − (−1/2)) = 5/(3/2) = 10/3|r| < 1, so use a/(1 − r)

    ✓ 10/3 ≈ 3.333333; the partial sums 5, 2.5, 3.75, 3.125 … zig-zag in on it.

  4. Find Σk = 0∞ 3(0.2)k. Geometric

    Show answer and steps
    a = 3, r = 0.2First term (k = 0) and ratio3/(1 − 0.2) = 3.75Answer

    ✓ 3 + 0.6 + 0.12 + 0.024 = 3.744 after four terms.

  5. Does Σn = 1∞ n/(n + 1) converge? nth-term test

    Show answer and steps
    n/(n + 1) → 1Limit of the terms1 ≠ 0, so the series divergesnth-term test

    ✓ You are adding numbers close to 1 for ever, so the sums grow past any bound.

  6. Does Σn = 1∞ 3n/n! converge? Ratio test

    Show answer and steps
    an+1/an = 3/(n + 1)Ratio of consecutive terms→ 0 < 1, so it convergesRatio test

    ✓ Its sum is e³ − 1 ≈ 19.085537, from the series for eˣ.

  7. Does Σn = 1∞ n²/2n converge? Ratio test

    Show answer and steps
    an+1/an = ((n + 1)/n)² · 1/2Ratio of consecutive terms→ 1/2 < 1, so it convergesRatio test

    ✓ Its sum is 6; the first ten terms already add to about 5.857.

  8. Write the degree-4 Taylor polynomial of cos x about 0 and use it to estimate cos 0.5. Taylor

    Show answer and steps
    P₄(x) = 1 − x²/2 + x⁴/24Even powers onlyP₄(0.5) = 1 − 0.125 + 0.0026041667Substitute x = 0.5≈ 0.877604Estimate

    ✓ cos 0.5 ≈ 0.877583, so the error is about 0.000022.

  9. How many terms of 1 + 1/2 + 1/3 + … are needed for the sum to pass 3? Harmonic

    Show answer and steps
    S₁₀ ≈ 2.928968Ten terms: not yetS₁₁ ≈ 3.019877Eleven terms: past 3

    ✓ Passing 10 takes 12,367 terms. The sums grow without bound, but very slowly.

Section XIISelf-guided quiz: eight quick questions

Choose one answer for each question, then press “Check my answers.” Your best score is saved only in this browser.

1. What is the limit of the sequence an = n/(n + 1)?
2. What is the sum of 1 + 1/3 + 1/9 + 1/27 + …?
3. The harmonic series 1 + 1/2 + 1/3 + 1/4 + …
4. If the terms aₙ go to 0, the series Σ aₙ
5. For a series of positive terms, the ratios aₙ₊₁/aₙ tend to 1/3. The series
6. The degree-3 Taylor polynomial of eˣ about 0 is
7. The Taylor series of ln(1 + x) about 0 converges for
8. Who proved that the harmonic series diverges, around 1350?

Section XIIIFrom Zeno to Maclaurin

Infinite sums puzzled people long before calculus. In the fifth century BC, Zeno of Elea posed paradoxes about motion, recorded by Aristotle in his Physics. In the Dichotomy, a runner must first reach the halfway point, then half of what is left, and so on without end; in the Achilles, the swift Achilles can never overtake a slower runner who has a head start. Hidden inside is the sum 1/2 + 1/4 + 1/8 + …, which we now know is exactly 1.

Around 1350 the French scholar Nicole Oresme proved that the harmonic series diverges, in his Questiones super geometriam Euclidis, by grouping the terms into blocks of 2, 4, 8, … terms, the same proof as in Section V. His Questiones circulated in manuscript and were not printed until the twentieth century.

In Kerala, in south India, Madhava of Sangamagrama found series equivalent to those for sin x, cos x and arctan x around 1400, more than two hundred years before they were found again in Europe. We know these results mainly through the works of later Kerala mathematicians, such as Jyesthadeva, whose Yukti-Bhasa, written in Malayalam around 1550, explains how the series were found.

In England, Brook Taylor described what we call Taylor’s theorem in a letter of 26 July 1712, and published it in his 1715 book Methodus incrementorum directa et inversa. He was not alone: James Gregory, Newton, Leibniz, Johann Bernoulli and de Moivre had all found versions of it. Its importance was not widely seen until 1772, when Lagrange called it the basic principle of the differential calculus.

The Scottish mathematician Colin Maclaurin used the special case centred at 0 in his two-volume Treatise of Fluxions (1742), and it is now called the Maclaurin series. Maclaurin did not claim it as his own discovery: he acknowledged Taylor’s work.

NotesSources and credits

Credits: every graph and diagram on this page is original SVG, computed from the actual functions, made for Jugami. The explorers run live in your browser with plain JavaScript, no outside libraries.