Corrected exercises: sequences (MATH 141)

Read the pattern in three layers. The signs alternate and the first term is positive: $(-1)^{n+1}$. The numerators $3, 4, 5, 6, 7$ are $n + 2$. The denominators $5, 25, 125, 625, 3125$ are $5^n$. So $a_n = (-1)^{n+1}\frac{n+2}{5^n}$; check with $n = 4$: $(-1)^5 \cdot \frac{6}{625} = -\frac{6}{625}$, as listed. For the limit, the sign prevents any direct computation, so pass to the absolute value: $|a_n| = \frac{n+2}{5^n}$. With $f(x) = \frac{x+2}{5^x}$, of the form $\frac{\infty}{\infty}$, L'Hôpital's rule gives $\lim_{x\to\infty}\frac{1}{5^x \ln 5} = 0$, and since $f(n) = |a_n|$, $|a_n| \to 0$. By the theorem, $a_n \to 0$. Writing $(-1)^n$ with the wrong starting sign costs the formula mark: always test the formula on the first two listed terms.

Divide numerator and denominator by $n^2$, the dominant power: $\frac{3 - \frac{1}{n} + \frac{4}{n^2}}{5 + \frac{2}{n}} \to \frac{3}{5}$, since $\frac{1}{n} \to 0$ and $\frac{1}{n^2} \to 0$. For the second, the dominant power under the root is $n^2$, so divide by $n = \sqrt{n^2}$, legitimate because $n > 0$: $\frac{2 + \frac{1}{n}}{\sqrt{9 + \frac{1}{n}}} \to \frac{2}{\sqrt 9} = \frac{2}{3}$. The trap is to divide the numerator by $n$ and the root by $n$ as well, giving $\sqrt{\frac{9n^2 + n}{n}}$: the $n$ goes inside the root as $n^2$.

The even terms are $c_{2k} = \frac{2k}{2k+1} \to 1$ and the odd terms are $c_{2k+1} = -\frac{2k+1}{2k+2} \to -1$ as $k \to \infty$, which is exactly what the figure shows: two clouds of dots settling on the two dashed lines. If $\{c_n\}$ converged to some $L$, every subsequence would converge to the same $L$, so we would have $L = 1$ and $L = -1$ at once. Contradiction: $\{c_n\}$ diverges. Note that the sequence is bounded, $|c_n| < 1$: bounded is not enough to converge.

$|d_n| = \frac{1}{\sqrt n} \to 0$, so $d_n \to 0$ by the theorem: the sign alternates, but the terms are crushed toward $0$ from both sides. In c), $|c_n| = \frac{n}{n+1} \to 1$, and the theorem does not apply, because it is a statement about the limit $0$ ONLY. From $|a_n| \to 1$ one can conclude nothing: $1, 1, 1, \dots$ converges to $1$, while $c_n$ diverges. Invoking the absolute value theorem with a limit other than $0$ is a lost mark on every paper that does it.

$\infty - \infty$ is an indeterminate form, not $0$: both terms grow, and the question is which one grows faster, by how much. Multiply by the conjugate: $\sqrt{n^2 + 3n} - n = \frac{(n^2 + 3n) - n^2}{\sqrt{n^2 + 3n} + n} = \frac{3n}{\sqrt{n^2 + 3n} + n}$. Divide by $n$: $\frac{3}{\sqrt{1 + \frac{3}{n}} + 1} \to \frac{3}{1 + 1} = \frac{3}{2}$. Sanity check with $n = 100$: $\sqrt{10300} \approx 101.49$, and $101.49 - 100 = 1.49$, close to $1.5$. The difference of two quantities that tend to infinity can tend to any number, to infinity, or to nothing at all.

The figure does NOT give the answer: the dots rise to about $0.74$ near $n = 7$ and are still above $0.5$ at $n = 40$, so the first forty terms suggest nothing close to $0$. The tail decides, and the tail needs a computation. Let $f(x) = \frac{\ln x}{\sqrt x}$ for $x \ge 1$. As $x \to \infty$ this is of the form $\frac{\infty}{\infty}$, and $f$ is differentiable, so L'Hôpital's rule applies: $\lim_{x\to\infty} \frac{1/x}{1/(2\sqrt x)} = \lim_{x\to\infty} \frac{2}{\sqrt x} = 0$. Since $a_n = f(n)$ for every $n$, $\lim a_n = 0$. The expected write-up has exactly these three moves: define $f$, apply the rule to $f$, come back to $a_n$ by the theorem. Writing $\lim \frac{\ln n}{\sqrt n} = \lim \frac{1/n}{1/(2\sqrt n)}$ differentiates with respect to an integer, and a marker takes the method mark away for it.

$f(x) = \frac{x^2}{e^x}$ is of the form $\frac{\infty}{\infty}$. L'Hôpital once: $\frac{2x}{e^x}$, still $\frac{\infty}{\infty}$; twice: $\frac{2}{e^x} \to 0$. Hence $\frac{n^2}{e^n} \to 0$. Each application must be announced with its form: the rule is only licensed while the form is still $\frac{\infty}{\infty}$ or $\frac{0}{0}$, and a student who applies it once more to $\frac{2}{e^x}$ obtains $\frac{0}{e^x}$, a correct number by luck and a wrong method.

A variable exponent calls for the logarithm. Let $c_n = n^{1/n}$, so $\ln c_n = \frac{\ln n}{n}$. With $f(x) = \frac{\ln x}{x}$, L'Hôpital gives $\frac{1/x}{1} \to 0$, so $\ln c_n \to 0$. The exponential is continuous at $0$, so $c_n = e^{\ln c_n} \to e^0 = 1$. The last sentence is the one students forget: finding $\ln c_n \to 0$ and answering $0$ gives the limit of the LOGARITHM, not of the sequence. For $(n^2 + 1)^{1/n}$, the logarithm is $\frac{\ln(n^2 + 1)}{n}$; L'Hôpital on $\frac{\ln(x^2+1)}{x}$ gives $\frac{2x}{x^2 + 1} \to 0$, so the limit is again $e^0 = 1$. The base tends to infinity, the exponent tends to $0$: this form $\infty^0$ is indeterminate, and here the exponent wins.

$n \sin\frac{1}{n}$ is of the form $\infty \cdot 0$. Rewrite it as a quotient with $t = \frac{1}{x}$, so $t \to 0^+$ as $x \to \infty$: $x \sin\frac{1}{x} = \frac{\sin t}{t} \to 1$, the limit of MATH 140 (or L'Hôpital: $\frac{\cos t}{1} \to 1$). So $n \sin\frac{1}{n} \to 1$. For the second, the same substitution gives $\frac{1 - \cos t}{t^2}$, of the form $\frac{0}{0}$; L'Hôpital gives $\frac{\sin t}{2t} \to \frac{1}{2}$. So $n^2\left(1 - \cos\frac{1}{n}\right) \to \frac{1}{2}$. The substitution $t = \frac{1}{x}$ turns a limit at infinity into a limit at $0$, where the standard limits live.

The claim is FALSE. For every integer $n$, $\sin(\pi n) = 0$, so $\{a_n\}$ is the constant sequence $0, 0, 0, \dots$ and converges to $0$. The function $\sin(\pi x)$ does oscillate between $-1$ and $1$ forever, but the integers land exactly on its zeros. The theorem goes in ONE direction: if $f(x) \to L$ then $f(n) \to L$. The converse is false, so the divergence of $f$ proves nothing about the sequence. In practice: when $f(x)$ has a limit, use it; when it has none, go back to the sequence itself.

Since $-1 \le \cos n \le 1$, we have $1 \le 2 + \cos n \le 3$, and dividing by $n > 0$ keeps the inequalities: $\frac{1}{n} \le \frac{2 + \cos n}{n} \le \frac{3}{n}$. Both bounds tend to $0$, so by the squeeze theorem $\frac{2 + \cos n}{n} \to 0$. The figure shows the whole mechanism: the dots wander because $\cos n$ does, but they are trapped in a funnel whose two walls both go to $0$. Never try to compute $\lim \cos n$: it does not exist, and the squeeze theorem is exactly the tool that avoids needing it.

$\frac{n!}{n^n} = \frac{1 \cdot 2 \cdot 3 \cdots n}{n \cdot n \cdot n \cdots n} = \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n}{n}$. Each of the fractions $\frac{2}{n}, \dots, \frac{n}{n}$ lies in $(0, 1]$, so their product is at most $1$, and $0 < \frac{n!}{n^n} \le \frac{1}{n}$. Both bounds tend to $0$: $\frac{n!}{n^n} \to 0$. L'Hôpital is powerless here, since $n!$ is not a function of a real variable in MATH 141; writing the quotient as a product is THE gesture for factorials.

Factor out the dominant term: $2^n + 7^n$ lies between $7^n$ and $7^n + 7^n = 2 \cdot 7^n$. Taking $n$-th roots, which preserves inequalities between positive numbers: $7 \le (2^n + 7^n)^{1/n} \le 2^{1/n} \cdot 7$. Now $2^{1/n} = e^{(\ln 2)/n} \to e^0 = 1$, so the upper bound tends to $7$, as does the lower bound. The limit is $7$. The smaller base disappears completely: the $n$-th root of a sum of exponentials tends to the LARGEST base.

By definition, $\sqrt n - 1 < \lfloor \sqrt n \rfloor \le \sqrt n$. Dividing by $\sqrt n > 0$: $1 - \frac{1}{\sqrt n} < \frac{\lfloor \sqrt n \rfloor}{\sqrt n} \le 1$. Both bounds tend to $1$, so the limit is $1$. The floor function jumps, and its sequence is not monotone ($\frac{\lfloor \sqrt 8 \rfloor}{\sqrt 8} = \frac{2}{\sqrt 8} \approx 0.71$ while $\frac{\lfloor \sqrt 9 \rfloor}{\sqrt 9} = 1$), but the jumps shrink relative to $\sqrt n$, and the squeeze sees only that.

In the first argument the bounds are $-1$ and $1$: they converge to two DIFFERENT limits, and the squeeze theorem then says nothing, which is just as well since that sequence diverges (Exercise 1 c). The second argument draws a conclusion the theorem does not contain: the squeeze gives a limit only when both bounds share it, and it never says that a limit exists and lies between two bounds. Choose better bounds instead: $0 \le \sin^2 n \le 1$ gives $0 \le \frac{\sin^2 n}{n} \le \frac{1}{n}$, and both tend to $0$, so $\frac{\sin^2 n}{n} \to 0$. The rule to remember: a bound that does not tend to the target is not a bound, it is a waste of a line.

$\left|-\frac{3}{4}\right| < 1$, so $\left(-\frac{3}{4}\right)^n \to 0$: the sign alternates, the size shrinks. $\frac{5}{4} > 1$, so $\left(\frac{5}{4}\right)^n \to \infty$. $(-1)^n$ takes the values $-1, 1, -1, \dots$: its even and odd terms have different limits, so it diverges. $\left(-\frac{5}{4}\right)^n$ diverges too, since $\left|\left(-\frac{5}{4}\right)^n\right| \to \infty$ while the sign alternates: it tends neither to $\infty$ nor to $-\infty$, and the correct word is diverges, not tends to infinity. The case $r = -1$ is the one students misplace: it is inside $[-1, 1]$ and yet the sequence diverges.

Divide by the dominant exponential, the one with the largest base. In the first, it is $3^n$: $\frac{\left(\frac{2}{3}\right)^n + 3}{1 - \frac{4}{3^n}} \to \frac{0 + 3}{1 - 0} = 3$, because $3^{n+1} = 3 \cdot 3^n$. In the second, it is $5^n$: numerator $5^n - 16 \cdot 4^n$, denominator $5 \cdot 5^n + 3^n$, so $\frac{1 - 16\left(\frac{4}{5}\right)^n}{5 + \left(\frac{3}{5}\right)^n} \to \frac{1}{5}$. The shifted exponents $3^{n+1}$ and $4^{n+2}$ are only constants in disguise, $3 \cdot 3^n$ and $16 \cdot 4^n$: rewrite them BEFORE dividing.

$\frac{a_{n+1}}{a_n} = \frac{10^{n+1}}{(n+1)!} \cdot \frac{n!}{10^n} = \frac{10}{n+1}$. This quotient is larger than $1$ for $n \le 8$, equal to $1$ for $n = 9$ and smaller than $1$ for $n \ge 10$: the sequence climbs until $a_9 = a_{10} = \frac{10^9}{9!}$, larger than $2700$, then decreases, which is the hump of the figure. For $n \ge 10$, every new factor $\frac{10}{k}$ with $k \ge 11$ is at most $\frac{10}{11}$, so $0 < a_n \le a_{10}\left(\frac{10}{11}\right)^{n-10}$. Since $\left|\frac{10}{11}\right| < 1$, the upper bound tends to $0$, and so does $a_n$ by the squeeze theorem. Looking at the first nine terms and concluding that the sequence tends to infinity is the error the figure is designed to provoke: the tail decides.

With $f(x) = \frac{x^3}{2^x}$, of the form $\frac{\infty}{\infty}$, L'Hôpital three times gives $\frac{3x^2}{2^x \ln 2}$, $\frac{6x}{2^x (\ln 2)^2}$, then $\frac{6}{2^x (\ln 2)^3} \to 0$, so $\frac{n^3}{2^n} \to 0$. For $\frac{2^n}{n!}$ the method of c) applies: $\frac{2^n}{n!} = \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdots \frac{2}{n} \le 2 \cdot 1 \cdot \left(\frac{2}{3}\right)^{n-2}$ for $n \ge 2$, and $\left(\frac{2}{3}\right)^{n-2} \to 0$. The two rungs of the ladder need two different tools: L'Hôpital for a power against an exponential, a product and a squeeze as soon as a factorial appears.

Take the logarithm: $\ln a_n = 100 \ln n - n \ln 1.01 = n\left(\frac{100 \ln n}{n} - \ln 1.01\right)$. By Exercise 2, $\frac{\ln n}{n} \to 0$, so the bracket tends to $-\ln 1.01$, a NEGATIVE number since $1.01 > 1$, and $\ln a_n \to -\infty$. Hence $a_n = e^{\ln a_n} \to 0$. The terms are enormous for a very long time: $a_{1000} = \frac{10^{300}}{1.01^{1000}}$ and $1.01^{1000} = e^{1000 \ln 1.01} < e^{10} < 3^{10} < 10^5$, because $\ln 1.01 < 0.01$ (the graph of $\ln(1 + u)$ lies below its tangent line $y = u$), so $a_{1000} > 10^{295}$. Yet the exponential wins in the end. Any exponential with base $b > 1$, however close to $1$, beats any power of $n$.