11. 수열의 극한Limits of Sequences

Proof.

(1) Let $\epsilon \gt 0$ be given. By the Archimedean property, there exists a natural number $N$ with $N \gt 1/\epsilon$. Then for $n \geq N$, $$\left|\frac{1}{n} - 0\right| = \frac{1}{n} \leq \frac{1}{N} \lt \epsilon,$$ so $\displaystyle\lim_{n \to \infty} \frac{1}{n} = 0$.

(2) Let $\epsilon \gt 0$ be given. Take $N = 1$ (since $k$ is constant, any $N$ works). Then for $n \geq N$, $|k - k| = 0 \lt \epsilon$, so $\displaystyle\lim_{n \to \infty} k = k$.

Proof. Suppose the sequence $a_n = (-1)^n$ converges to some real number $L$.

Take $\epsilon = 1$. By the definition of convergence, there exists a natural number $N$ such that $|a_n - L| \lt 1$ for all $n \geq N$. Since $n = 2N$ is even, $a_{2N} = 1$, and since $n = 2N+1$ is odd, $a_{2N+1} = -1$. Both indices satisfy $n \geq N$, so $$|1 - L| \lt 1, \qquad |-1 - L| \lt 1$$ must both hold. The first inequality gives $0 \lt L \lt 2$, and the second gives $-2 \lt L \lt 0$. No $L$ satisfies both conditions at once, a contradiction.

Hence $a_n = (-1)^n$ does not converge, i.e., it diverges.

Proof. For every natural number $n$, $$\big||a_n| - 0\big| = |a_n| = |a_n - 0|,$$ so for any $\epsilon \gt 0$, the condition $|a_n - 0| \lt \epsilon$ and the condition $\big||a_n| - 0\big| \lt \epsilon$ are exactly the same condition.

Hence the statement "for every $\epsilon \gt 0$ there is a natural number $N$ such that $|a_n - 0| \lt \epsilon$ for $n \geq N$" (i.e., $\displaystyle\lim_{n\to\infty} a_n = 0$) and the statement "for every $\epsilon \gt 0$ there is a natural number $N$ such that $\big||a_n| - 0\big| \lt \epsilon$ for $n \geq N$" (i.e., $\displaystyle\lim_{n\to\infty} |a_n| = 0$) are equivalent.

Proof.

(1) Here $c_{2k-1} = a_k$ and $c_{2k} = b_k$ for $k = 1, 2, \cdots$.

($\Rightarrow$) Suppose $\displaystyle\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = L$. Let $\epsilon \gt 0$ be given. There exist $N_1$ such that $n \geq N_1 \Rightarrow |a_n - L| \lt \epsilon$, and $N_2$ such that $n \geq N_2 \Rightarrow |b_n - L| \lt \epsilon$. Let $M = \max\{N_1, N_2\}$ and take $N = 2M - 1$. For $m \geq N$: if $m = 2k-1$ (odd), then $k \geq M \geq N_1$, so $|c_m - L| = |a_k - L| \lt \epsilon$; if $m = 2k$ (even), then $2k \geq 2M - 1$ gives $k \geq M \geq N_2$, so $|c_m - L| = |b_k - L| \lt \epsilon$. Hence $m \geq N \Rightarrow |c_m - L| \lt \epsilon$, so $\displaystyle\lim_{n\to\infty} c_n = L$.

($\Leftarrow$) Suppose $\displaystyle\lim_{n\to\infty} c_n = L$. Let $\epsilon \gt 0$ be given. There exists $N$ such that $m \geq N \Rightarrow |c_m - L| \lt \epsilon$. Take $N_0 = N$. For $n \geq N_0$, we have $2n - 1 \geq n \geq N$ and $2n \geq n \geq N$, so $|a_n - L| = |c_{2n-1} - L| \lt \epsilon$ and $|b_n - L| = |c_{2n} - L| \lt \epsilon$. Hence $\displaystyle\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = L$.

(2) Suppose $\{a_n\}$ converges to $L$. Since $\{n_k\}$ is an increasing sequence of natural numbers, induction gives $n_k \geq k$ for every $k$ (indeed $n_1 \geq 1$, and $n_{k+1} \gt n_k \geq k$ implies $n_{k+1} \geq k+1$).

Let $\epsilon \gt 0$ be given. There exists a natural number $N$ such that $n \geq N \Rightarrow |a_n - L| \lt \epsilon$. If $k \geq N$, then $n_k \geq k \geq N$, so $|a_{n_k} - L| \lt \epsilon$. Hence $\displaystyle\lim_{k\to\infty} a_{n_k} = L$, i.e., the subsequence $\{a_{n_k}\}$ also converges.

Solution.

(1) $f(x) = \dfrac{\ln x}{x}$ is continuous for $x \geq 1$, and by L'Hôpital's Rule, $$\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0,$$ so $\displaystyle\lim_{n \to \infty} \frac{\ln n}{n} = 0$.

(2) Substituting $m = n - 1$, $$\left(\frac{n+1}{n-1}\right)^n = \left(1 + \frac{2}{n-1}\right)^n = \left(1 + \frac{2}{m}\right)^{m+1} = \left(1 + \frac{2}{m}\right)^m \left(1 + \frac{2}{m}\right),$$ and as $m \to \infty$, $\left(1+\frac{2}{m}\right)^m \to e^2$ and $\left(1+\frac{2}{m}\right) \to 1$, so the limit is $e^2$.

(3) Write $\sqrt[n]{n} = 1 + b_n$ with $b_n \geq 0$. By the binomial theorem, for $n \geq 2$, $$n = (1+b_n)^n \geq \binom{n}{2} b_n^2 = \frac{n(n-1)}{2} b_n^2,$$ so $b_n^2 \leq \dfrac{2}{n-1} \to 0$. Hence $b_n \to 0$ and $\sqrt[n]{n} = 1 + b_n \to 1$.

(4) The case $x = 1$ is trivial. If $x \gt 1$, write $x^{1/n} = 1 + c_n$ with $c_n \gt 0$; by Bernoulli's inequality, $$x = (1+c_n)^n \geq 1 + nc_n,$$ so $0 \lt c_n \leq \dfrac{x-1}{n} \to 0$, i.e., $x^{1/n} \to 1$. If $0 \lt x \lt 1$, then $y = 1/x \gt 1$ gives $y^{1/n} \to 1$, so $x^{1/n} = 1/y^{1/n} \to 1$.

(5) Let $a_n = \dfrac{|x|^n}{n!}$. The case $x = 0$ is trivial. For $x \neq 0$, $$\frac{a_{n+1}}{a_n} = \frac{|x|}{n+1} \to 0,$$ so for some natural number $N \gt 2|x|$, $n \geq N$ implies $\dfrac{a_{n+1}}{a_n} \leq \dfrac{1}{2}$, hence $0 \leq a_n \leq a_N \left(\dfrac12\right)^{n-N} \to 0$. That is, $\displaystyle\lim_{n\to\infty}\frac{x^n}{n!} = 0$.

(6) The case $x = 0$ is trivial. For $x \neq 0$, for $n$ sufficiently large $1 + x/n \gt 0$, so $$\left(1+\frac{x}{n}\right)^n = \exp\left(n\ln\left(1+\frac{x}{n}\right)\right) = \exp\left(x \cdot \frac{\ln(1+x/n)}{x/n}\right).$$ Since $u = x/n \to 0$ and $\displaystyle\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$ (by L'Hôpital's Rule), the exponent converges to $x \cdot 1 = x$, and by continuity of $\exp$, $$\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n = e^x.$$

Solution. We use the results from Example 02: $\displaystyle\lim_{n\to\infty}\frac{\ln n}{n}=0$, $\displaystyle\lim_{n\to\infty}\sqrt[n]{n}=1$, $\displaystyle\lim_{n\to\infty}\frac{x^n}{n!}=0$, and $\displaystyle\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = e^x$.

(1) $\dfrac{\ln(n^2)}{n} = \dfrac{2\ln n}{n} \to 2 \cdot 0 = 0$.

(2) $\sqrt[n]{n^2} = (\sqrt[n]{n})^2 \to 1^2 = 1$.

(3) $\sqrt[n]{3n} = 3^{1/n}\sqrt[n]{n}$, and $3^{1/n} = e^{(\ln 3)/n} \to e^0 = 1$, so $\sqrt[n]{3n} \to 1 \cdot 1 = 1$.

(4) Since $\left|-\dfrac12\right| \lt 1$, we get $\left(-\dfrac12\right)^n \to 0$.

(5) $\left(\dfrac{n-2}{n}\right)^n = \left(1+\dfrac{-2}{n}\right)^n \to e^{-2}$.

(6) $\dfrac{100^n}{n!} \to 0$.

(7) For $n \geq 1$, $0 \lt \dfrac{n!}{n^n} = \dfrac1n \cdot \dfrac2n \cdots \dfrac{n}{n} \leq \dfrac1n$, and $\dfrac1n \to 0$, so by the Squeeze Theorem $\dfrac{n!}{n^n} \to 0$.

(8) For $n \geq 1$, $n \leq n^2+n \leq 2n^2$, so $\sqrt[n]{n} \leq \sqrt[n]{n^2+n} \leq \sqrt[n]{2n^2} = 2^{1/n}(\sqrt[n]{n})^2$. As $n \to \infty$ both bounds converge to $1$, so by the Squeeze Theorem $\sqrt[n]{n^2+n} \to 1$.

(9) $\tan^{-1} n$ is bounded, $0 \leq \tan^{-1} n \lt \dfrac{\pi}{2}$, and $\dfrac{1}{\sqrt n} \to 0$, so $\dfrac{1}{\sqrt n}\tan^{-1} n \to 0$.

Proof. Since $f(0) = 0$, $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h}.$$ That is, for every $\epsilon \gt 0$ there exists $\delta \gt 0$ such that $0 \lt |h| \lt \delta$ implies $$\left|\frac{f(h)}{h} - f'(0)\right| \lt \epsilon.$$ Choose a natural number $N \gt 1/\delta$. Then for $n \geq N$, $h = 1/n$ satisfies $0 \lt h \lt \delta$, so $$\left|\frac{f(1/n)}{1/n} - f'(0)\right| = |nf(1/n) - f'(0)| \lt \epsilon.$$ Hence $\displaystyle\lim_{n\to\infty} a_n = \lim_{n\to\infty} nf(1/n) = f'(0)$.

Let $f(x) = \tan^{-1} x$. Then $f(0) = 0$ and $f'(x) = \dfrac{1}{1+x^2}$, so $f'(0) = 1$. By the result above, $$\lim_{n \to \infty} n\tan^{-1}\frac{1}{n} = f'(0) = 1.$$

Proof.

(1) Let $\epsilon \gt 0$ be given. By the Archimedean property, there exists a natural number $N$ with $N \gt \epsilon^{-1/c}$. Since $x \mapsto x^c$ ($x \gt 0$) is increasing for $c \gt 0$, $n \geq N$ gives $$n^c \geq N^c \gt \epsilon^{-1},$$ and hence $$\left|\frac{1}{n^c} - 0\right| = \frac{1}{n^c} \lt \epsilon.$$ That is, $\displaystyle\lim_{n\to\infty}\frac{1}{n^c} = 0$.

(2) $f(x) = \dfrac{\ln x}{x^c}$ is continuous for $x \geq 1$, and by L'Hôpital's Rule, $$\lim_{x \to \infty} \frac{\ln x}{x^c} = \lim_{x \to \infty} \frac{1/x}{cx^{c-1}} = \lim_{x \to \infty} \frac{1}{cx^c} = 0$$ (since $c \gt 0$, $x^c \to \infty$). Since this continuous-variable limit is $0$, the corresponding sequence limit obtained by substituting $x = n$ is also $0$, i.e., $$\lim_{n \to \infty} \frac{\ln n}{n^c} = 0.$$

Solution.

(1) Since $a_n = n$ grows without bound as $n \to \infty$, no real number $M$ can satisfy $a_n \leq M$ for all $n$ (by the Archimedean property). Hence the set of upper bounds is the empty set $\varnothing$, and no supremum exists.

On the other hand, if $M \leq 1$ then $a_n = n \geq 1 \geq M$ for all $n \geq 1$, so $M$ is a lower bound; if $M \gt 1$ then $a_1 = 1 \lt M$, so $M$ is not a lower bound. Hence the set of lower bounds is $(-\infty, 1]$, and the infimum is $1$, which is in fact attained as the minimum at $a_1 = 1$.

(2) Since $a_{n+1} - a_n = (n+1) - n = 1 \gt 0$, the sequence $a_n$ is monotonically increasing.

Solution.

(1) Since $a_n = \dfrac{n}{n+1} = 1 - \dfrac{1}{n+1} \lt 1$, if $M \geq 1$ then $a_n \lt 1 \leq M$ for all $n$, so $M$ is an upper bound. If $M \lt 1$, then since $a_n \to 1$, some $n$ satisfies $a_n \gt M$, so $M$ is not an upper bound. Hence the set of upper bounds is $[1, \infty)$, and the supremum is $1$, which is not attained (there is no maximum).

On the other hand, as shown in (2), $a_n$ is increasing, so its minimum is $a_1 = \dfrac12$. If $M \leq \dfrac12$ then $a_n \geq a_1 = \dfrac12 \geq M$ for all $n$, so $M$ is a lower bound; if $M \gt \dfrac12$ then $a_1 = \dfrac12 \lt M$, so $M$ is not a lower bound. Hence the set of lower bounds is $\left(-\infty, \dfrac12\right]$, and the infimum is $\dfrac12$, attained as the minimum at $a_1$.

(2) For $n \geq 1$, $$a_{n+1} - a_n = \frac{n+1}{n+2} - \frac{n}{n+1} = \frac{(n+1)^2 - n(n+2)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)} \gt 0,$$ so $a_n$ is monotonically increasing.

Proof. We show $0 \leq a_n \leq 4$ for all $n$ by induction. Since $a_1 = 0$, the base case holds. Suppose $a_n \leq 4$; then $$a_{n+1} = \sqrt{8+2a_n} \leq \sqrt{8+8} = 4,$$ and $a_{n+1} = \sqrt{8+2a_n} \geq 0$, so $0 \leq a_{n+1} \leq 4$. Hence $0 \leq a_n \leq 4$ for all $n$.

Next we show $a_n$ is monotonically increasing by induction. Since $a_2 = \sqrt{8} \gt 0 = a_1$, the base case holds. Suppose $a_{n+1} \geq a_n$; since the square root function is increasing, $$a_{n+2} = \sqrt{8+2a_{n+1}} \geq \sqrt{8+2a_n} = a_{n+1}.$$ Hence $a_n$ is monotonically increasing.

Since $a_n$ is increasing and bounded above, by the Monotone Convergence Theorem it converges to some $L$. Taking the limit of both sides of the recurrence (using continuity of the square root), $$L = \sqrt{8+2L},$$ and squaring both sides, $$L^2 = 8+2L \quad\Longleftrightarrow\quad L^2-2L-8=0 \quad\Longleftrightarrow\quad (L-4)(L+2)=0,$$ so $L=4$ or $L=-2$. Since $a_n \geq 0$, we have $L \geq 0$, so $L=4$.

Proof. We show $0 \leq a_n \leq 1$ for all $n$ by induction. Since $a_1 = 0$, the base case holds. Suppose $0 \leq a_n \leq 1$; then $2+a_n \gt 0$ and $$a_{n+1} - 1 = \frac{1+2a_n}{2+a_n} - 1 = \frac{a_n - 1}{2+a_n} \leq 0,$$ so $a_{n+1} \leq 1$, and since $1+2a_n \geq 0$ and $2+a_n \gt 0$, also $a_{n+1} \geq 0$. Hence $0 \leq a_n \leq 1$ for all $n$.

Next, $$a_{n+1} - a_n = \frac{1+2a_n}{2+a_n} - a_n = \frac{1+2a_n - a_n(2+a_n)}{2+a_n} = \frac{1-a_n^2}{2+a_n},$$ and since $0 \leq a_n \leq 1$, we have $1-a_n^2 \geq 0$ and $2+a_n \gt 0$. Hence $a_{n+1} - a_n \geq 0$, so $a_n$ is monotonically increasing.

Since $a_n$ is increasing and bounded above, by the Monotone Convergence Theorem it converges to some $L$. Taking the limit of both sides of the recurrence, $$L = \frac{1+2L}{2+L} \quad\Longleftrightarrow\quad L(2+L) = 1+2L \quad\Longleftrightarrow\quad L^2 = 1,$$ so $L = \pm 1$. Since $0 \leq a_n \leq 1$, we have $L \geq 0$, so $L=1$.

Proof. Let $f(x) = x^2-2$, so $f'(x) = 2x$. Newton's method gives the recurrence $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^2-2}{2x_n} = \frac{x_n^2+2}{2x_n} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right).$$ Since $x_0 = 1 \gt 0$, induction shows $x_n \gt 0$ for all $n$.

Also, $$x_{n+1} - \sqrt2 = \frac{x_n^2+2}{2x_n} - \sqrt2 = \frac{x_n^2 - 2\sqrt2\, x_n + 2}{2x_n} = \frac{(x_n-\sqrt2)^2}{2x_n} \geq 0,$$ so $x_n \geq \sqrt2$ for $n \geq 1$. (Indeed $x_1 = \dfrac12(1+2) = \dfrac32 \geq \sqrt2$.)

For $n \geq 1$, $$x_{n+1} - x_n = \frac{1}{2}\left(x_n+\frac{2}{x_n}\right) - x_n = \frac{2-x_n^2}{2x_n},$$ and since $x_n \geq \sqrt2$ gives $x_n^2 \geq 2$ and $x_n \gt 0$, we get $x_{n+1} - x_n \leq 0$. Hence $\{x_n\}_{n \geq 1}$ is monotonically decreasing.

Since $\{x_n\}_{n \geq 1}$ is decreasing and bounded below (by $\sqrt2$), the Monotone Convergence Theorem shows it converges. (Convergence is unaffected by dropping finitely many initial terms, so $\{x_n\}$ as a whole converges too.) In fact the limit $L$ must satisfy $L = \dfrac12\left(L+\dfrac2L\right)$, so $L = \sqrt2$.