PG4.2 Limits of sums and products

Simulate

Check the claim with numbers

Lesson PG4.1 proved that $a_n = \frac{3n + 1}{n + 2} \to 3$ . Take a second sequence, $b_n = 1 + \frac{(-1)^n}{n}$ , which tends to 1 from both sides. If limits behave like numbers, the sum should tend to $3 + 1 = 4$ and the product to $3 \cdot 1 = 3$ .

from fractions import Fraction def a(n): return Fraction(3 * n + 1, n + 2) def b(n): return 1 + Fraction((-1) ** n, n) for n in [10, 100, 1000]: print(n, round(float(a(n)), 6), round(float(b(n)), 6), round(float(a(n) + b(n)), 6), round(float(a(n) * b(n)), 6))

⌘/Ctrl + Enter runs the cell

At $n = 1000$ the sum is 3.99601 and the product 2.998005, close to 4 and 3. The next cell tries the idea a proof will use. To get the sum within 0.01 of 4, ask each sequence to be within half of that, 0.005, of its own limit.

from fractions import Fraction def a(n): return Fraction(3 * n + 1, n + 2) def b(n): return 1 + Fraction((-1) ** n, n) eps = Fraction(1, 100) N1 = max(n for n in range(1, 20001) if abs(a(n) - 3) >= eps / 2) N2 = max(n for n in range(1, 20001) if abs(b(n) - 1) >= eps / 2) N = max(N1, N2) print(N1, N2, N) print(all(abs(a(n) + b(n) - 4) < eps for n in range(N + 1, N + 10001))) print(max(n for n in range(1, 20001) if abs(a(n) + b(n) - 4) >= eps))

⌘/Ctrl + Enter runs the cell

$a_n$ is within 0.005 of 3 after $n = 998$ , and $b_n$ within 0.005 of 1 after $n = 200$ . After the larger of the two, 998, the sum is within 0.01 of 4: the next 10,000 terms all pass. The sum's own last miss is at 597, so halving the tolerance is generous, and that is fine.

⌘/Ctrl + Enter runs the cell

The rules need both limits to exist. Neither $(-1)^n$ nor $(-1)^{n+1}$ has a limit, yet their sum is 0 for every $n$ . And $n$ grows without end while $\frac{1}{n} \to 0$ , yet their product is 1 every time.

Conjecture

a_n \to a

The numbers fit, for these two sequences and one tolerance. The proof must work for every pair of convergent sequences and every $\varepsilon > 0$ .

The triangle inequality

The proofs measure a sum of two distances. One inequality controls it: the size of a sum is at most the sum of the sizes.

Rule: Triangle inequality

|x + y| \le |x| + |y|

It holds for all real numbers $x$ and $y$ . Equality holds when $x$ and $y$ have the same sign, and the left side is smaller when the signs differ.

Here is why. By the definition of absolute value (lesson 0.3.3), $-|x| \le x \le |x|$ and $-|y| \le y \le |y|$ . Adding gives $-(|x| + |y|) \le x + y \le |x| + |y|$ , which says exactly that $|x + y| \le |x| + |y|$ .

x = 3

⌘/Ctrl + Enter runs the cell

Splitting the tolerance

For the sum, the distance from $a + b$ is $(a_n - a) + (b_n - b)$ , so by the triangle inequality it is at most $|a_n - a| + |b_n - b|$ . Make each part less than $\frac{\varepsilon}{2}$ , and the sum is less than $\varepsilon$ .

Each sequence has its own $N$ : $N_1$ for $a_n$ and $N_2$ for $b_n$ . After the larger of the two, both distances are small at once, as the simulation found with 998 and 200.

Format 1

Reorder the steps

Here is the proof of the sum rule, with its lines shuffled. Put them in order with the arrows, or drag a line by its handle.

By the triangle inequality, $|(a_n + b_n) - (a + b)| \le |a_n - a| + |b_n - b| < \varepsilon$ .
Let $\varepsilon > 0$ . Then $\frac{\varepsilon}{2} > 0$ is a tolerance too.
Let $N$ be the larger of $N_1$ and $N_2$ , and take any $n > N$ .
Since $a_n \to a$ and $b_n \to b$ , there are $N_1$ and $N_2$ with $|a_n - a| < \frac{\varepsilon}{2}$ for $n > N_1$ and $|b_n - b| < \frac{\varepsilon}{2}$ for $n > N_2$ .
This holds for every $n > N$ , and $\varepsilon > 0$ was arbitrary, so $a_n + b_n \to a + b$ .
Then $n > N_1$ and $n > N_2$ , so both distances are less than $\frac{\varepsilon}{2}$ .

The product rule

For the product, add and subtract $a_n b$ to split the difference into two parts:

\begin{gathered} a_n b_n - ab \\ = a_n(b_n - b) + b(a_n - a) \end{gathered}

The second part is $|b|$ times a small distance. The first is $|a_n|$ times a small distance, so it needs a bound on $|a_n|$ that does not grow with $n$ .

Format 2

Fill the gap

The proof of the product rule, with its key steps left out. Type each missing piece as you would a quick answer, such as a(y-x) or 0.25, then check.

|a_n|

How to type answers

Powers with ^, brackets round a power that is a sum: r^k, r^(k+1).
Fractions with / and brackets: (1-r)/(1+r). Multiply with * or side by side: 2k+1.
Press Enter in any box to check them all.

Format 3

Find the error

Another proof of the product rule. One line states something false about convergent sequences. Select it.

Simulate

A claim of your own to prove

Multiplying every term by a fixed number $c$ should multiply the limit by $c$ . The cell scales $a_n = \frac{3n + 1}{n + 2}$ by $c = 5$ , $-2$ and 0, and finds the last term more than 0.01 from $3c$ .

from fractions import Fraction def a(n): return Fraction(3 * n + 1, n + 2) eps = Fraction(1, 100) for c in [5, -2, 0]: outside = [n for n in range(1, 20001) if abs(c * a(n) - 3 * c) >= eps] print(c, max(outside) if outside else "none")

⌘/Ctrl + Enter runs the cell

The last miss is at $n = 2498$ for $c = 5$ and at 998 for $c = -2$ : a bigger $|c|$ needs a later start. For $c = 0$ no term misses at all, because every term is 0.

Conjecture

a_n \to a

Three values of $c$ fit. The proof must cover every $c$ , and the simulation hints that $c = 0$ needs its own case.

Format 4

Write it

Prove from the definition that if $a_n \to a$ , then $c\,a_n \to c\,a$ for every number $c$ . Name your $N$ explicitly, and watch for $c = 0$ .

Your proof

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Write your own first: a proof you have tried to write teaches far more than one you only read.

With $c = -1$ and the sum rule, the difference rule follows: $a_n - b_n \to a - b$ .

|x + y| \le |x| + |y|

Limits of sums and products

Check the claim with numbers

The triangle inequality

Rule: Triangle inequality

Splitting the tolerance

Reorder the steps

The product rule

Fill the gap

Find the error

A claim of your own to prove

Write it

Summary card

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