Sequences & series: practice problems with step-by-step solutions

Also called: infinite series · convergence tests · Taylor series · power series

Free Sequences & series practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: arithmetic and geometric series, convergence and power series, Taylor and Maclaurin polynomials.

Free practice problemEasy
kak​12345678

Find the sum S8 of the first 8 terms of the geometric sequence 5, 7.5, 11.25, …

a₁ = 5r = 1.5n = 8
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What you practise

  1. Easyarithmetic and geometric series
  2. Mediumconvergence and power series
  3. HardTaylor and Maclaurin polynomials

How to solve Sequences & series problems

  1. Identify the type: arithmetic (constant difference), geometric (constant ratio), a standard series, a power series or a Taylor polynomial.
  2. For sums use Sₙ = n(a₁ + aₙ)/2, Sₙ = a₁(1 − rⁿ)/(1 − r), or S = a₁/(1 − r) when |r| < 1, with a₁ the first term of the sum.
  3. For convergence: n-th term test first, then compare with Σ1/kᵖ or a geometric series, and use the ratio or root test for factorials and k-th powers.
  4. For Taylor polynomials list f(a), f′(a), f″(a), …, divide the k-th derivative by k!, and bound the error with M·|x − a|ⁿ⁺¹/(n + 1)!.

Common mistakes

  • Using n instead of n − 1 in aₙ = a₁ + (n − 1)d or a₁rⁿ⁻¹.
  • Using the infinite-sum formula when |r| ≥ 1, or with the wrong first term.
  • Concluding convergence only because the terms tend to 0.
  • Inverting the ratio-test limit: R = lim |cₖ/cₖ₊₁|, not its reciprocal.
  • Forgetting the factorials or the alternating signs in a Taylor polynomial.

Key formulas

Arithmetic sequence
aₙ = a₁ + (n − 1)d, Sₙ = n(a₁ + aₙ)/2
Geometric sequence
aₙ = a₁rⁿ⁻¹, Sₙ = a₁(1 − rⁿ)/(1 − r)
Infinite geometric series
S = a₁/(1 − r), |r| < 1
p-series
Σ1/kᵖ converges only for p > 1
Ratio test
L = lim |ak+1/ak|: L < 1 converges, L > 1 diverges
Radius of convergence
R = lim |ck/ck+1|
Taylor polynomial
Pₙ(x) = Σ f⁽ᵏ⁾(a)(x − a)ᵏ/k!
Maclaurin series
eˣ = Σxᵏ/k!, sin x = x − x³/3! + …, cos x = 1 − x²/2! + …, ln(1 + x) = x − x²/2 + …
Lagrange error bound
|Rₙ| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!

Learn Sequences & series from scratch

Before this: Limits, Derivatives

1Arithmetic and geometric sequences

A sequence is a list of numbers a₁, a₂, a₃, … and a series is their sum. Two kinds come back in every exam.

An arithmetic sequence adds the same difference d each step: 3, 7, 11, 15, … has d = 4.

aₙ = a₁ + (n − 1)d Sₙ = n(a₁ + aₙ)/2

For 3, 7, 11, …: a₂₀ = 3 + 19·4 = 79 and S₂₀ = 20·(3 + 79)/2 = 820. Note the n − 1: from the first to the 20th term there are only 19 steps.

A geometric sequence multiplies by the same ratio r each step: 2, 6, 18, … has r = 3.

aₙ = a₁·rⁿ⁻¹ Sₙ = a₁(1 − rⁿ)/(1 − r)
Self-check What is the 10th term of 5, 8, 11, 14, …?
  1. 29
  2. 30
  3. 32
  4. 35

Answer: 32. d = 3, so a₁₀ = 5 + 9·3 = 32. Using 10·3 instead of 9·3 gives 35.

2Infinite geometric series

Keep adding terms of 2 + 1 + 0.5 + 0.25 + … and the sum creeps up to 4 without ever passing it.

nSₙ4123456789233.5

When |r| < 1 the term rⁿ in Sₙ shrinks to 0, and the infinite sum exists:

S = a₁/(1 − r), only for |r| < 1

Here a₁ is the first term of the sum. For Σk=2∞ 8·(1/2)ᵏ the first term is 8·(1/2)² = 2, so S = 2/(1 − 1/2) = 4.

When |r| ≥ 1 the terms do not shrink and the series has no sum: it diverges.

Self-check What is Σk=0∞ 3·(1/4)ᵏ?
  1. 12
  2. 2.4
  3. 3
  4. 4

Answer: 4. The first term (k = 0) is 3 and r = 1/4, so S = 3/(1 − 1/4) = 4. Using 1 + r in the denominator gives 2.4.

3Does a series converge?

Most series have no simple sum formula, but you can still decide whether the sum is finite. Use the tests in this order:

  1. n-th term test: if the terms do not tend to 0, the series diverges. Σ k/(k + 1) diverges because its terms tend to 1.
  2. Standard series: Σ1/kᵖ converges only for p > 1 (so Σ1/k, the harmonic series, diverges). Σrᵏ converges only for |r| < 1.
  3. Comparison: 1/(k² + 3) is smaller than 1/k², so Σ1/(k² + 3) converges too.
  4. Ratio test: L = lim |ak+1/ak|. L < 1 converges, L > 1 diverges, L = 1 gives no answer. Use it for factorials and k-th powers.
  5. Alternating series: Σ(−1)ᵏ/k converges, because the terms alternate in sign and decrease to 0.

Example: for Σ k/2ᵏ the ratio is ((k + 1)/k)·(1/2), which tends to L = 1/2 < 1, so it converges.

Self-check Which of these series diverges?
  1. Σ (2/3)ᵏ
  2. Σ 1/k²
  3. Σ 1/√k
  4. Σ 3ᵏ/k!

Answer: Σ 1/√k. 1/√k = 1/k1/2 is a p-series with p = 1/2 ≤ 1, so it diverges. 1/k² has p = 2, (2/3)ᵏ is geometric with r < 1, and the ratio test gives L = 0 for 3ᵏ/k!.

4Power series and the radius of convergence

A power series Σ ck(x − x₀)ᵏ is a polynomial without an end. It converges for x close enough to the centre x₀.

The ratio test on the whole term gives the radius of convergence R: the series converges for |x − x₀| < R and diverges for |x − x₀| > R.

R = lim |ck/ck+1|

Example: Σ xᵏ/(k·3ᵏ) has ck/ck+1 = 3(k + 1)/k, which tends to 3. So R = 3 and the series converges for −3 < x < 3.

At the two end points the test gives no answer; check them separately if the question asks for the full interval.

5Taylor and Maclaurin polynomials

Near a point a, a smooth function looks like a polynomial. The Taylor polynomial of degree n matches the value and the first n derivatives at a:

Pₙ(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n!

For a = 0 it is called the Maclaurin polynomial. Learn these series by heart:

eˣ = 1 + x + x²/2! + x³/3! + …
sin x = x − x³/3! + x⁵/5! − …
cos x = 1 − x²/2! + x⁴/4! − …
ln(1 + x) = x − x²/2 + x³/3 − …
xeˣP₁P₂1

The more terms you take, the longer the polynomial stays close to the function.

Worked example

Approximate e0.5 with the Maclaurin polynomial of degree 3, and bound the error.

1

Write the polynomial

P₃(x) = 1 + x + x²/2 + x³/6
2

Substitute x = 0.5

P₃(0.5) = 1 + 0.5 + 0.125 + 0.020833 = 1.645833
3

Lagrange error bound

The error is at most M·|x|⁴/4!, where M is the largest value of the 4th derivative eᵗ between 0 and 0.5: M = e0.5 ≈ 1.6487.

|R₃| ≤ 1.6487·0.5⁴/24 = 0.0043
4

Compare

The calculator gives e0.5 = 1.648721, so the true error is 0.0029, inside the bound.

Answer

e0.5 ≈ 1.6458, with an error of at most 0.0043.

6Common mistakes and exam questions

  • Using n instead of n − 1 steps for the n-th term.
  • Writing a/(1 − r) with the wrong first term when the sum does not start at k = 0.
  • Concluding that a series converges because its terms tend to 0. Σ1/k is the counterexample.
  • Forgetting the factorials in a Taylor polynomial, or the minus signs of sin and cos.
  • Using |x|ⁿ/n! instead of |x|ⁿ⁺¹/(n + 1)! in the error bound.

Easy: sums of arithmetic and geometric sequences. Medium: deciding convergence, the ratio test and the radius of convergence. Hard: Taylor polynomials, their coefficients and error bounds.

Self-check What is the coefficient of x³ in the Maclaurin series of sin x?
  1. 1/3
  2. 1/6
  3. −1/3
  4. −1/6

Answer: −1/6. sin x = x − x³/3! + …, and 3! = 6, so the coefficient is −1/6. Forgetting the minus sign gives 1/6.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
kak​13579111315ε = 0.2

The geometric sequence aₙ = 5·0.85ⁿ⁻¹ decreases. Find the first index n for which aₙ < 0.2.

a₁ = 5r = 0.85ε = 0.2

Find n

Show hints
  1. Solve 5·0.85ⁿ⁻¹ < 0.2 with logarithms.
  2. ln 0.85 is negative: dividing by it turns the inequality around.
  3. n − 1 > ln(0.2/5)/ln 0.85 = 19.81
Show the answer and the step-by-step solution

Arithmetic and geometric series: an arithmetic sequence adds the same difference d each step, aₙ = a₁ + (n − 1)d, and its sum is Sₙ = n(a₁ + aₙ)/2. A geometric sequence multiplies by the same ratio r, aₙ = a₁·rⁿ⁻¹, with Sₙ = a₁(1 − rⁿ)/(1 − r). When |r| < 1 the terms shrink fast enough for the infinite sum to exist: S = a₁/(1 − r), where a₁ is the first term of the sum.

Inequality

5·0.85ⁿ⁻¹ < 0.2 → 0.85ⁿ⁻¹ < 0.04

Logarithms

(n − 1)·ln 0.85 < ln 0.04 → n − 1 > 19.81
n = 21

Check

a20 = 0.227997 ≥ 0.2
a21 = 0.193798 < 0.2

Answer

n = 21
Example 2Medium
kSn​12345678910S = ?

Find the sum S = Σk=1∞ 12/(k(k + 3)).

ak = 12/(k(k + 3))

Find S

Show hints
  1. Partial fractions: 1/(k(k + 3)) = (1/3)·(1/k − 1/(k + 3)).
  2. Write out the first few terms: most of them cancel (a telescoping sum).
  3. What remains is (12/3)·(1 + … + 1/3).
Show the answer and the step-by-step solution

Convergence: if the terms do not tend to 0 the series diverges (n-th term test). Σ1/kᵖ converges only for p > 1, and Σrᵏ only for |r| < 1. Compare a series with one of these when its terms behave like them for large k. Ratio test: L = lim |ak+1/ak|; root test: L = lim |ak|1/k. L < 1 converges, L > 1 diverges, L = 1 decides nothing. A power series Σck(x − x₀)ᵏ converges for |x − x₀| < R with R = lim |ck/ck+1|; an alternating series with terms decreasing to 0 converges.

Partial fractions

12/(k(k + 3)) = (12/3)·(1/k − 1/(k + 3))

Cancel

S = (12/3)·(1 + 1/2 + 1/3) = 7.333

Answer

S = 7.333
Example 3Hard

f(x) = cos x is approximated at x = 1 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 1 × 10⁻⁶.

f(x) = cos xx = 1M = 1ε = 1 × 10⁻⁶

Find n

Show hints
  1. Try n = 1, 2, 3, … and compute 1·1ⁿ⁺¹/(n + 1)! each time.
  2. The factorial grows fast, so a few steps are enough.
  3. Stop at the first n where the bound is below 1 × 10⁻⁶.
Show the answer and the step-by-step solution

Taylor polynomials: near x = a a smooth function is approximated by Pₙ(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n!; for a = 0 this is the Maclaurin polynomial. Standard series: eˣ = Σxᵏ/k!, sin x = x − x³/3! + x⁵/5! − …, cos x = 1 − x²/2! + x⁴/4! − …, ln(1 + x) = x − x²/2 + x³/3 − …, 1/(1 − x) = Σxᵏ. Lagrange error bound: |f(x) − Pₙ(x)| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!, where M is the largest value of |f⁽ⁿ⁺¹⁾| between a and x.

n(n + 1)!bound011120.5260.16673240.0416741200.00833357200.001389650400.00019847403202.48 × 10⁻⁵83628802.756 × 10⁻⁶936288002.756 × 10⁻⁷

Bounds for increasing n

n = 8: 2.756 × 10⁻⁶ ≥ 1 × 10⁻⁶
n = 9: 2.756 × 10⁻⁷ < 1 × 10⁻⁶

Smallest degree

n = 9

Answer

n = 9

Questions and answers

What is Sequences & series?

A sequence is a list of numbers a₁, a₂, a₃, … and a series is their sum. Two kinds come back in every exam.

An arithmetic sequence adds the same difference d each step: 3, 7, 11, 15, … has d = 4.

How do I solve Sequences & series problems step by step?
  1. Identify the type: arithmetic (constant difference), geometric (constant ratio), a standard series, a power series or a Taylor polynomial.
  2. For sums use Sₙ = n(a₁ + aₙ)/2, Sₙ = a₁(1 − rⁿ)/(1 − r), or S = a₁/(1 − r) when |r| < 1, with a₁ the first term of the sum.
  3. For convergence: n-th term test first, then compare with Σ1/kᵖ or a geometric series, and use the ratio or root test for factorials and k-th powers.
  4. For Taylor polynomials list f(a), f′(a), f″(a), …, divide the k-th derivative by k!, and bound the error with M·|x − a|ⁿ⁺¹/(n + 1)!.
What are common mistakes in Sequences & series?
  • Using n instead of n − 1 in aₙ = a₁ + (n − 1)d or a₁rⁿ⁻¹.
  • Using the infinite-sum formula when |r| ≥ 1, or with the wrong first term.
  • Concluding convergence only because the terms tend to 0.
  • Inverting the ratio-test limit: R = lim |cₖ/cₖ₊₁|, not its reciprocal.
  • Forgetting the factorials or the alternating signs in a Taylor polynomial.
What do the Sequences & series exercises cover?
  • Easy: arithmetic and geometric series
  • Medium: convergence and power series
  • Hard: Taylor and Maclaurin polynomials

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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