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For every $\varepsilon$ an $N$: simplifying the distance, solving for the index, and why one close term proves nothing.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to prove that a sequence converges straight from the definition — fixing an arbitrary epsilon, simplifying the distance to the limit, solving for the index and naming an N that works for the whole tail. You will also be able to negate the definition correctly, which is what a proof of divergence needs, and to say why a term that is occasionally close proves nothing.
The Archimedean property, which turns a threshold into a whole number, and the triangle inequality, which bounds a quantity by the pieces you control. A convergence proof is those two used in sequence, and nothing else.
A sequence is a function from $\mathbb{N}$ to $\mathbb{R}$, written $(a_n)$. It converges to $L$, written $a_n \to L$, when the condition below holds; $L$ is its limit. A sequence that converges to no number diverges — and a divergent sequence may be perfectly well behaved, like $(-1)^n$, or may run off to infinity.
> $a_n \to L$ means: for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $|a_n - L| < \varepsilon$ for all $n \ge N$.
Read it as a challenge and a response. The challenge is $\varepsilon$, an accuracy demanded by somebody else. The response is $N$, an index beyond which the sequence meets it. The sequence converges when there is a response to every challenge.
Three features of the order matter.
$\varepsilon$ comes first, so $N$ may depend on it. Almost always it must: a smaller $\varepsilon$ needs a larger $N$. An $N$ that did not depend on $\varepsilon$ would say the terms are eventually exactly $L$.
$N$ comes before $n$, so $N$ may not depend on $n$. This is the commonest error. One term close to $L$ is nothing; the definition asks for a tail.
The inequality is strict and the tail is infinite. Finitely many terms may be anywhere at all. Changing the first thousand terms of a sequence changes neither whether it converges nor what to.
A convergence proof is not an observation that the terms get close. It is a rule that turns a given $\varepsilon$ into an $N$, and the rule is the proof: no $N$, no theorem.
Another way: picture
Draw a horizontal band of half-width $\varepsilon$ about the height $L$. Convergence says: whatever the band's width, all but finitely many of the dots lie inside it. Narrow the band and more dots fall outside — but always only finitely many, and $N$ is the index after which none do.
Another way: steps
To prove $a_n \to L$ from the definition:
Step 2 is where the work is; steps 1, 4 and 5 are the same every time.
Claim: $\dfrac{3n + 1}{n + 2} \to 3$.
Simplify. $\left|\dfrac{3n+1}{n+2} - 3\right| = \left|\dfrac{3n + 1 - 3n - 6}{n+2}\right| = \dfrac{5}{n+2}$.
Solve. $\dfrac{5}{n+2} < \varepsilon$ exactly when $n > \dfrac{5}{\varepsilon} - 2$.
Name $N$. Let $N$ be any whole number past $5/\varepsilon$. The Archimedean property says there is one.
Verify. For $n \ge N$, $\dfrac{5}{n+2} \le \dfrac{5}{N+2} < \dfrac{5}{N} < \varepsilon$.
Notice the slack taken in the last line. $N$ was not chosen to be the smallest index that works, only an index that works, and the estimate was made cruder twice to keep the algebra simple. A convergence proof does not have to be sharp. Any $N$ that does the job proves the theorem, and insisting on the least one is a different and harder question with no mathematical reward.
A limit is unique. If $a_n \to L$ and $a_n \to M$, then for any $\varepsilon$, both $|L - a_n|$ and $|a_n - M|$ are eventually below $\varepsilon/2$, so $|L - M| < \varepsilon$ by the triangle inequality. A non-negative number below every positive number is $0$.
A convergent sequence is bounded. Take $\varepsilon = 1$. Beyond $N$ the terms lie in $(L-1, L+1)$; before $N$ there are finitely many, so take the largest modulus among them. The bound is the maximum of the two.
Both proofs use a specific $\varepsilon$ — a half share, then $1$ — and that is the characteristic move of the subject. The definition is a supply of estimates, and a proof consists of asking it for the right one.
Choosing $N$ after $n$. 'Given $n$, choose $N = n$' is not a response to the challenge; it is a restatement of it. $N$ is named once per $\varepsilon$ and then every later term must comply.
Thinking the terms must approach monotonically. $(-1)^n/n$ converges to $0$ while jumping from side to side. The definition says nothing about direction, only about distance.
Reading 'gets close to' as the definition. $(-1)^n$ gets close to $1$ infinitely often and converges to nothing. Close infinitely often is not the same as close from some point on, and that distinction is exactly what $N$ encodes.
Claim: $\dfrac{n}{2n + 5} \to \dfrac{1}{2}$. Let $\varepsilon > 0$.
The challenge is fixed first.
Simplify: $\left|\dfrac{n}{2n+5} - \dfrac{1}{2}\right| = \dfrac{5}{2(2n+5)} < \dfrac{5}{4n}$.
Crude is fine, so long as it is an upper bound.
Choose $N$ past $5/(4\varepsilon)$. Then for $n \ge N$ the distance is below $\varepsilon$, so the limit is $1/2$.
The response, and the check for the whole tail.
Claim: $(-1)^n$ converges to no limit $L$. Take $\varepsilon = 1$.
Divergence needs one epsilon, not all of them.
Whatever $N$ is offered, the terms at $N$ and $N+1$ are $1$ and $-1$ in some order, and they are $2$ apart.
Find terms beyond every $N$ that stay apart.
They cannot both be within $1$ of $L$, since that would put them within $2$ of each other with room to spare. So no $L$ works.
The triangle inequality, doing the contradiction.
The numerator never exceeds $1$ in size, so the distance to $0$ is at most $1/n$.
Bound the part you cannot compute.
Given $\varepsilon$, choose $N$ past $1/\varepsilon$; then $1/n \le 1/N < \varepsilon$ for all $n \ge N$.
The same response as always.
So the limit is $0$. Nothing was needed about the behaviour of $\sin n$, which is chaotic — only that it is bounded, which is the general lesson: a bounded quantity multiplied by one that tends to zero tends to zero.
Put the steps of the proof that $a_n = 6 + \dfrac{1}{n}$ converges to $6$ into the order they must be made in.
Number the steps in order (write the number in the box):
$a_n = \dfrac{1}{n} \to 0$. For $\varepsilon = \dfrac{1}{6}$, what is the smallest $N$ with $|a_n - 0| < \varepsilon$ for every $n \ge N$?
Answer:
$a_n = \dfrac{2n + 1}{n + 3}$ converges to $2$. Give the distance $|a_n - 2|$ at each index.
| distance from the limit | |
|---|---|
| index $4$ | |
| index $6$ | |
| index $9$ |
Match each sequence to what it does.
| converges | bounded, but has no limit | unbounded | |
|---|---|---|---|
| $a_n = 7/n$ | |||
| $a_n = (-1)^n 7$ | |||
| $a_n = 7n$ | |||
| $a_n = 7 + 1/n$ |
Here is an argument that $a_n = 4/n$ converges to $0$. Every line is true, and one of them does not establish what it claims. Mark it.
This task has no paper form; do it on a device.
To show that $a_n = (-1)^n 9$ does not converge to $0$, which statement has to be established?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of the proof that $a_n = 8 + \dfrac{1}{n}$ converges to $8$ into the order they must be made in.
Number the steps in order (write the number in the box):
You can produce the N that a given epsilon needs, simplify the distance to a limit, and negate the definition to prove divergence. Say in your own words why N is allowed to depend on epsilon but not on n. Next: computing limits without going back to the definition every time.
10. Your turn: prove that $\dfrac{\sin n}{n} \to 0$, step 3