Limit of a sequence from the definition, squeeze and bounds¶

Math I: Calculus · Week 2 test, open question · 7 points · Lecture 3

Problem¶

Define, for $n\geq1\text{,}$

a_n=\frac{2(-1)^n}{n^2+n},\quad b_n=\sqrt{2}\cos\left(\frac{1}{n}\right).

(a) Prove, using the definition, that $(a_n)_{n=1}^{\infty}$ converges to $0\text{.}$ [2 points]
(b) Determine whether the sequence $a_1b_1,a_2b_2,\dots$ is convergent. If it is, find its limit; if not, justify. [2 points]
(c) Evaluate $\inf_{n\geq1}a_n-\sup_{n\geq1}b_n$ and briefly justify both bounds. [3 points]

What you need¶

Limit of a sequence: the $\varepsilon$ – $N$ definition.
Squeeze theorem for sequences.
Bounded sequence: a bounded sequence times a sequence tending to $0$ tends to $0\text{.}$
Infimum and supremum, and the difference with minimum and maximum.
Trigonometric functions: $|\cos x|\leq1\text{,}$ and $\cos x<1$ for $0<x\leq1\text{.}$

Plan¶

In (a), find a simple upper bound for $|a_n|$ that clearly goes to $0\text{,}$ and choose $N$ from it. In (b), do not compute $\lim b_n$ at all: boundedness of $(b_n)$ is enough. In (c), list the first few terms to see where the extreme values sit, then justify each bound separately. One of them is attained and the other is not.

Step-by-step solution¶

Try each step yourself before you open it.

(a) Step 1 · Bound the distance to the limit and choose N (1 point)

Let $\varepsilon>0\text{.}$ Since $n^2+n>n\text{,}$

|a_n-0|=\frac{2}{n^2+n}<\frac{2}{n}.

So it is enough to make $2/n<\varepsilon\text{.}$ Choose a positive integer $N>2/\varepsilon\text{.}$

Concept: Limit of a sequence

(a) Step 2 · Verify for all n ≥ N (1 point)

If $n\geq N\text{,}$ then

|a_n-0|<\frac{2}{n}\leq\frac{2}{N}<\varepsilon.

Since $\varepsilon>0$ was arbitrary, $a_n\to0$ by definition.

(b) Step 1 · Bound the second factor (1 point)

Because $|\cos x|\leq1$ for all $x\text{,}$ we have $|b_n|\leq\sqrt{2}$ for every $n\text{.}$ You do not need the limit of $(b_n)$ here.

(b) Step 2 · Squeeze (1 point)

Then

0\leq|a_nb_n|\leq\sqrt{2}\,|a_n|\to0.

By the squeeze theorem $|a_nb_n|\to0\text{,}$ so $a_nb_n\to0\text{.}$ The sequence converges, with limit $0\text{.}$

Concept: Squeeze theorem for sequences

(c) Step 1 · The infimum of (a_n) (1 point)

The sign of $a_n$ alternates: $a_n>0$ for even $n$ and $a_n<0$ for odd $n\text{.}$ For odd $n$ the size $2/(n^2+n)$ decreases as $n$ grows:

a_1=-1,\quad a_3=-\frac{1}{6},\quad a_5=-\frac{1}{15},\ \dots

So every term is at least $a_1=-1\text{,}$ and $-1$ is itself a term. Hence $\inf_{n\geq1}a_n=-1\text{;}$ it is even a minimum.

(c) Step 2 · The supremum of (b_n) (1 point)

For every $n\text{,}$ $0<1/n\leq1\text{,}$ so $\cos(1/n)<1$ and $b_n<\sqrt{2}\text{.}$ Thus $\sqrt{2}$ is an upper bound.

It is the least upper bound because $1/n\to0$ and cosine is continuous, so $b_n\to\sqrt{2}\text{.}$ For any $\eta>0$ some $b_n$ exceeds $\sqrt{2}-\eta\text{,}$ so no smaller number is an upper bound. Hence $\sup_{n\geq1}b_n=\sqrt{2}\text{,}$ although no term equals $\sqrt{2}\text{.}$

Concept: Supremum

(c) Step 3 · Combine (1 point)

\inf_{n\geq1}a_n-\sup_{n\geq1}b_n=-1-\sqrt{2}.

Where marks are lost¶

Choosing $N$ that depends on $n\text{.}$ In (a), $N$ may depend on $\varepsilon$ only.
Trying to compute $\lim b_n$ in (b) and multiplying limits. That works here, but boundedness is all you need, and the bounded-times-null argument also works when $(b_n)$ has no limit.
Saying "the supremum is the largest term". In (c) the supremum $\sqrt{2}$ is not attained. You must show it is an upper bound and that nothing smaller is.
Ignoring the sign of $a_n\text{.}$ The infimum comes from the negative terms, not from the limit $0\text{.}$

Try next¶

Continuity of $1/x^2$ from the $\varepsilon$ – $\delta$ definition: the same "bound, then choose" idea for functions.
Convergence of three series and $\displaystyle \sum\sin n$ : limits of sequences used to decide convergence of series.

Symbols: delta, epsilon, eta, infinity, square root, sum