The Lumenary · Grades 9–12

Sequences and series

Name: ______________________

Date: ______________________

You will learn to

Find formulas for the $n$ -th term and the partial sums of arithmetic and geometric sequences, and prove them by induction
Define sequences recursively and use summation notation
Explain why $1 + \tfrac12 + \tfrac14 + \cdots = 2$ and why $0.999\ldots = 1$ , and decide when an infinite geometric series converges
Use the term test, and prove that the harmonic series diverges although its terms tend to $0$

1.
Find the sum of the first $20$ terms of the arithmetic sequence $5, 8, 11, 14, \dots$
2.
Compute $3 + 6 + 12 + 24 + \dots + 3 \cdot 2^{9}$ (ten terms).
3.
A sequence is defined by $a_1 = 2$ and $a_{n+1} = 3a_n - 1$ . What is $a_5$ ?
4.
Which of these infinite series converge? (Choose all that apply.)
- A. $\displaystyle\sum_{k=0}^{\infty} \left(\tfrac{2}{3}\right)^k$
- B. $\displaystyle\sum_{k=0}^{\infty} \left(-\tfrac{9}{10}\right)^k$
- C. $\displaystyle\sum_{k=0}^{\infty} (-1)^k$
- D. $\displaystyle\sum_{k=1}^{\infty} \tfrac{1}{k}$
- E. $\displaystyle\sum_{k=1}^{\infty} \tfrac{1}{k^2}$
- F. $\displaystyle\sum_{k=0}^{\infty} \left(\tfrac{11}{10}\right)^k$
5.
Write $0.\overline{45} = 0.454545\ldots$ as a geometric series $a + ar + ar^2 + \cdots$ and find its sum $\tfrac pq$ in lowest terms. Enter $q$ .
6.
A student argues: "The terms $\tfrac{1}{\sqrt k}$ tend to $0$ , so by the term test $\sum_{k=1}^{\infty} \tfrac{1}{\sqrt k}$ converges." What is wrong?
- A.The term test only proves divergence; terms tending to $0$ prove nothing about convergence
- B.The terms $\tfrac{1}{\sqrt k}$ do not tend to $0$
- C.Nothing: the series converges
- D.The term test only applies to geometric series
7.
In the proof that the harmonic series diverges, why is $\tfrac{1}{9} + \tfrac{1}{10} + \dots + \tfrac{1}{16} \ge \tfrac12$ ?
- A.There are $8$ terms, each at least $\tfrac{1}{16}$
- B.There are $8$ terms, each at least $\tfrac{1}{9}$
- C.Because $\tfrac19 > \tfrac12$
- D.Because the terms tend to $0$
8.
In Lean, HasSum f S means "the series $\sum_{n \ge 0} f(n)$ converges to $S$ ". Which statement formalises $0.999\ldots = 1$ ?
- A.HasSum (fun n : ℕ => (9 : ℝ) / 10 ^ (n + 1)) 1
- B.HasSum (fun n : ℕ => (9 : ℝ) / 10 ^ n) 1
- C.(0.999 : ℝ) = 1
- D.∀ n : ℕ, ∑ k ∈ Finset.range n, (9 : ℝ) / 10 ^ (k + 1) = 1
9.
Write terms_needed(a, r, eps) for $0 < r < 1$ and $a > 0$ : the smallest $n$ such that the partial sum $S_n = a + ar + \dots + ar^{n-1}$ is within eps of the limit $\tfrac{a}{1 - r}$ . Use Fraction for exact arithmetic (the tests pass fractions). By Theorem 5 the gap is $\tfrac{a\,r^n}{1 - r}$ .
Computational — Practice

Answer key — Sequences and series

1.
$a_{20} = 5 + 19 \cdot 3 = 62$ , so $S_{20} = \tfrac{20(5 + 62)}{2} = 10 \cdot 67 = 670$ . Equivalently, $\tfrac{n(2a + (n - 1)d)}{2} = \tfrac{20(10 + 57)}{2} = 670$ .
2.
$\sum_{k=0}^{9} 3 \cdot 2^k = \tfrac{3(1 - 2^{10})}{1 - 2} = 3(1024 - 1) = 3069$ .
3.
$a_2 = 3 \cdot 2 - 1 = 5$ , $a_3 = 14$ , $a_4 = 41$ , $a_5 = 122$ . (In fact $a_n = \tfrac{3^n + 1}{2}$ , which you can prove by induction.)
4.
Converge: (a) sum $3$ , (b) sum $\tfrac{10}{19}$ , (e) sum $\tfrac{\pi^2}{6}$ . Diverge: (c) and (f) by the term test, and (d), the harmonic series, by Oresme's grouping — the term test says nothing about (d).
5.
$0.\overline{45} = \sum_{k=0}^{\infty} \tfrac{45}{100}\left(\tfrac{1}{100}\right)^k = \tfrac{45/100}{1 - 1/100} = \tfrac{45}{99} = \tfrac{5}{11}$ . So $q = 11$ .
6.
The term test says: convergent $\Rightarrow$ terms $\to 0$ . The student used the converse, which is false (the harmonic series is a counterexample). Here, since $\sqrt k \le k$ , each term $\tfrac{1}{\sqrt k} \ge \tfrac1k$ , so the partial sums are at least the harmonic partial sums, which are unbounded. The series diverges.
7.
The block $\tfrac19 + \dots + \tfrac1{16}$ has $16 - 8 = 8$ terms, and each is at least the smallest, $\tfrac{1}{16}$ . So the block is at least $8 \cdot \tfrac{1}{16} = \tfrac12$ . In general the block ending at $\tfrac{1}{2^{j+1}}$ has $2^j$ terms, each at least $\tfrac{1}{2^{j+1}}$ .
8.
$0.999\ldots = \sum_{n=0}^{\infty} \tfrac{9}{10^{n+1}}$ , which is option (a). Option (b) is off by one in the exponent, option (c) is a false statement about a finite decimal, and option (d) is false because each partial sum falls short of $1$ by $10^{-n}$ .
9.
```
from fractions import Fraction

def terms_needed(a, r, eps):
    n = 0
    while a * r ** n / (1 - r) >= eps:
        n += 1
    return n
```
For $1 + \tfrac12 + \tfrac14 + \cdots$ the gap after $n$ terms is $\tfrac{1}{2^{n-1}}$ , which first drops below $\tfrac{1}{1000}$ at $n = 11$ . With ratio $\tfrac{99}{100}$ the series converges much more slowly: $917$ terms are needed to get within $\tfrac{1}{100}$ . Convergence says the gap eventually drops below every $\varepsilon$ ; the loop finds when.