Exercises · 2. Sequences

Easy

1Easy

Let $a_n = \frac{n + \sin n}{n^2 + 1}$ . Prove directly from the $\varepsilon$ - $N$ definition that $\lim_{n \to \infty} a_n = 0$ , and give an explicit formula for a usable $N(\varepsilon)$ in terms of $\varepsilon > 0$ .

\left|\frac{n + \sin n}{n^2 + 1}\right| \le \frac{2}{n}

N(\varepsilon) = \left\lceil \frac{2}{\varepsilon} \right\rceil

The ε–N Definition

\varepsilon = 0.1

: the moment the tail fits entirely inside the

\varepsilon

-band is exactly the

N

we found.

Show solution

First trap the oscillating term in the numerator. Since $|\sin t| \le 1$ for every real $t$ , $|n + \sin n| \le n + 1$ . For $n \ge 1$ we have $n + 1 \le 2n$ and the denominator satisfies $n^2 + 1 \ge n^2$ , so
$|a_n - 0| = \frac{|n + \sin n|}{n^2 + 1} \le \frac{2n}{n^2} = \frac{2}{n}.$
Given $\varepsilon > 0$ , put $N(\varepsilon) = \lceil 2/\varepsilon \rceil$ . If $n > N$ then $n > 2/\varepsilon$ , i.e. $2/n < \varepsilon$ . Combined with the estimate above, $|a_n - 0| < \varepsilon$ , so $a_n \to 0$ by the $\varepsilon$ - $N$ definition.

2Easy

For $n \ge 1$ let $a_n = \frac{\lfloor n\sqrt{2} \rfloor}{n}$ , where $\lfloor x \rfloor$ is the greatest integer $\le x$ . (a) Show each $a_n$ is rational. (b) Prove $a_n \to \sqrt{2}$ from the $\varepsilon$ - $N$ definition. (c) Prove $\sqrt{2}$ is irrational. Conclude that a sequence of rationals can converge to an irrational limit.

0 \le n\sqrt{2} - \lfloor n\sqrt{2} \rfloor < 1

\left|a_n - \sqrt{2}\right| < \frac{1}{n}

The ε–N Definition

Show solution

(a) $\lfloor n\sqrt{2} \rfloor$ is an integer, so $a_n$ is an integer divided by $n$ : rational. (b) From the definition of the floor function, $0 \le n\sqrt{2} - \lfloor n\sqrt{2} \rfloor < 1$ . Dividing by $n$ gives $|a_n - \sqrt{2}| < 1/n$ . For $\varepsilon > 0$ put $N = \lceil 1/\varepsilon \rceil$ ; if $n > N$ then $|a_n - \sqrt{2}| < 1/n < \varepsilon$ , so $a_n \to \sqrt{2}$ .
(c) Suppose $\sqrt{2} = p/q$ with coprime positive integers $p, q$ . Then $p^2 = 2q^2$ , so $p^2$ is even, hence $p$ is even (an even square has an even root): $p = 2r$ . Substituting gives $4r^2 = 2q^2$ , i.e. $q^2 = 2r^2$ , so by the same argument $q$ is even. This contradicts coprimality. Hence $\sqrt{2}$ is irrational.
The rational sequence $(a_n)$ converges to the irrational $\sqrt{2}$ , so the expectation that a limit of rationals stays rational is false. A sequence limit can "escape" the set its terms live in.

3Easy

Let $a_n = (2 + (-1)^n)\left(1 - \frac{1}{n+1}\right)$ . Find $\limsup_{n \to \infty} a_n$ and $\liminf_{n \to \infty} a_n$ , with a rigorous proof using subsequence arguments.

a_{2k} = 3\left(1 - \frac{1}{2k+1}\right), \qquad a_{2k+1} = 1 - \frac{1}{2k+2}

Lim sup and Lim inf

Show solution

Separate even and odd terms. For $n = 2k$ , $(-1)^n = 1$ , so $a_{2k} = 3(1 - 1/(2k+1)) \to 3$ . For $n = 2k+1$ , $(-1)^n = -1$ , so $a_{2k+1} = 1 - 1/(2k+2) \to 1$ . Hence $\limsup a_n \ge 3$ and $\liminf a_n \le 1$ .
For the reverse inequalities, trap every term. Since $1 \le 2 + (-1)^n \le 3$ and $0 < 1 - 1/(n+1) < 1$ , we have $a_n < 3$ for all $n$ . Hence $\limsup a_n \le 3$ , and with the even subsequence, $\limsup a_n = 3$ . Likewise $a_n \ge 1 \cdot (1 - 1/(n+1))$ and $1 - 1/(n+1) \to 1$ , so $\liminf a_n \ge 1$ . With the odd subsequence, $\liminf a_n = 1$ .

Medium

4Medium

Let $a_n > 0$ with $a_n \to L > 0$ . Prove the geometric mean $G_n = (a_1 a_2 \cdots a_n)^{1/n}$ converges to $L$ . (Hint: taking logarithms turns this into a Cesaro mean.)

\log G_n = \frac{\log a_1 + \cdots + \log a_n}{n}

G_n = \exp\left(\frac{1}{n}\sum_{k=1}^{n} \log a_k\right) \to \exp(\log L) = L

Show solution

Put $b_n = \log a_n$ . Since $\log$ is continuous at $L > 0$ , $b_n \to \log L$ . By Cesaro's theorem (if a sequence converges, so do its arithmetic means, to the same limit),
$\frac{b_1 + \cdots + b_n}{n} \longrightarrow \log L \qquad (n \to \infty).$
But $\log G_n = \frac{1}{n}\sum_{k=1}^n \log a_k$ , so by continuity of the exponential, $G_n = \exp(\log G_n) \to \exp(\log L) = L$ . The hypothesis $L > 0$ is essential for taking logarithms.

Hard

5Hard

For a sequence with $a_n > 0$ , prove $\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \le \liminf_{n \to \infty} a_n^{1/n} \le \limsup_{n \to \infty} a_n^{1/n} \le \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}$ . Additionally, if $\frac{a_{n+1}}{a_n} \to L > 0$ , take logarithms and use the Stolz-Cesaro theorem to show $a_n^{1/n} \to L$ .

\liminf \frac{a_{n+1}}{a_n} \le \liminf a_n^{1/n} \le \limsup a_n^{1/n} \le \limsup \frac{a_{n+1}}{a_n}

\frac{a_{n+1}}{a_n} \to L > 0 \quad \Longrightarrow \quad a_n^{1/n} \to L

Show solution

The middle inequality $\liminf \le \limsup$ is trivial. We prove the last one. Let $R = \limsup a_{n+1}/a_n$ ; if $R = \infty$ there is nothing to prove, so assume $R < \infty$ . Take $\beta$ with $R < \beta < \infty$ . By definition, $a_{n+1}/a_n < \beta$ for $n \ge N$ . Hence $a_{N+k} < a_N \beta^k$ for $k \ge 1$ , and taking $(N+k)$ -th roots,
$a_{N+k}^{1/(N+k)} < a_N^{1/(N+k)} \cdot \beta^{k/(N+k)} \longrightarrow 1 \cdot \beta = \beta.$
( $a_N$ is fixed and the exponent tends to $0$ , so $a_N^{1/(N+k)} \to 1$ .) Hence $\limsup a_n^{1/n} \le \beta$ ; letting $\beta \downarrow R$ gives $\limsup a_n^{1/n} \le R$ . The first inequality is symmetric: let $r = \liminf a_{n+1}/a_n$ . If $r = 0$ there is nothing to prove; assume $r > 0$ . Take $0 < \alpha < r$ ; then $a_{n+1}/a_n > \alpha$ for $n \ge N$ , and from $a_{N+k} > a_N \alpha^k$ we get $a_{N+k}^{1/(N+k)} > a_N^{1/(N+k)} \alpha^{k/(N+k)} \to \alpha$ . So $\liminf a_n^{1/n} \ge \alpha$ , and $\alpha \uparrow r$ gives $\liminf a_n^{1/n} \ge r$ .
Finally the limit form: if $\frac{a_{n+1}}{a_n} \to L > 0$ then $\log a_{n+1} - \log a_n = \log\frac{a_{n+1}}{a_n} \to \log L$ . Applying Stolz-Cesaro to numerator $\log a_n$ and denominator $n$ gives $\frac{\log a_n}{n} \to \log L$ , and exponentiating, $a_n^{1/n} = \exp(\frac{\log a_n}{n}) \to \exp(\log L) = L$ . This is why the ratio test implies the root test.