Corrected exercises: the integral test and the comparison tests (MATH 141)

Let $f(x) = x e^{-x}$, so that $f(n) = n e^{-n} = a_n$. CONTINUOUS: a product of continuous functions. POSITIVE: $x > 0$ and $e^{-x} > 0$ on $[1, \infty)$. DECREASING: $f'(x) = e^{-x} - x e^{-x} = (1 - x)e^{-x}$, which is $\le 0$ for $x \ge 1$. The test applies from $N = 1$. By parts with $u = x$, $dv = e^{-x}\,dx$: $\int_1^t x e^{-x}\,dx = \left[-x e^{-x}\right]_1^t + \int_1^t e^{-x}\,dx = -t e^{-t} + e^{-1} - e^{-t} + e^{-1}$. As $t \to \infty$, $t e^{-t} = \frac{t}{e^t} \to 0$ by L'Hospital's rule and $e^{-t} \to 0$, so the integral converges to $\frac{2}{e}$. By the integral test, $\sum n e^{-n}$ CONVERGES. The conclusion names the test and both objects: the integral converges, therefore the series converges.

$f'(x) = \frac{\frac{1}{x}\cdot x - \ln x}{x^2} = \frac{1 - \ln x}{x^2}$. It is negative exactly when $\ln x > 1$, that is $x > e$: $f$ increases on $[1, e]$ and decreases on $[e, \infty)$, which is the peak of the figure. The hypothesis only has to hold on SOME $[N, \infty)$: take $N = 3 > e$. Look at the terms: $a_2 = \frac{\ln 2}{2}$ and $a_4 = \frac{\ln 4}{4} = \frac{2\ln 2}{4} = \frac{\ln 2}{2}$ are EQUAL, with $a_3$ larger, so the sequence is not decreasing from $n = 2$, and the test must start at $3$. Removing the finitely many terms $a_1 = 0$ and $a_2$ changes the sum but never the convergence. Now, with $u = \ln x$, $\int_3^t \frac{\ln x}{x}\,dx = \frac{(\ln t)^2}{2} - \frac{(\ln 3)^2}{2} \to \infty$. By the integral test $\sum_{n \ge 3} \frac{\ln n}{n}$ diverges, hence so does $\sum_{n \ge 1} \frac{\ln n}{n}$. The divergence test was useless here: $\frac{\ln n}{n} \to 0$.

$f(x) = \frac{\arctan x}{1 + x^2}$ is continuous and positive on $[1, \infty)$, since $\arctan x > 0$ for $x > 0$. The product of an INCREASING numerator and a decreasing $\frac{1}{1 + x^2}$ has no sign rule, so the derivative is needed: $f'(x) = \frac{\frac{1}{1 + x^2}(1 + x^2) - 2x\arctan x}{(1 + x^2)^2} = \frac{1 - 2x\arctan x}{(1 + x^2)^2}$. At $x = 1$ the numerator is $1 - 2 \cdot \frac{\pi}{4} = 1 - \frac{\pi}{2} < 0$, and $2x\arctan x$ increases, so the numerator stays negative on $[1, \infty)$: $f$ decreases there. With $u = \arctan x$, $du = \frac{dx}{1 + x^2}$: $\int_1^t f(x)\,dx = \frac{(\arctan t)^2}{2} - \frac{1}{2}\left(\frac{\pi}{4}\right)^2 \to \frac{1}{2}\cdot\frac{\pi^2}{4} - \frac{1}{2}\cdot\frac{\pi^2}{16} = \frac{3\pi^2}{32}$. The integral converges, so the series CONVERGES by the integral test.

The integral test gives the FATE of the series, never its value. Place over each $[n, n + 1]$ the rectangle of height $a_n = f(n)$: since $f$ is strictly decreasing on $[1, \infty)$, $f(x) < f(n)$ for $n < x \le n + 1$, so each rectangle has strictly more area than the region under the curve on the same interval. Adding them: $\sum_{n=1}^{\infty} a_n > \int_1^{\infty} f(x)\,dx = \frac{2}{e}$. The same picture shifted one step to the right, rectangles of height $a_{n+1}$ on $[n, n + 1]$, lies under the curve, so $\sum_{n=2}^{\infty} a_n < \int_1^{\infty} f = \frac{2}{e}$, and adding $a_1 = \frac{1}{e}$ gives $\sum_{n=1}^{\infty} a_n < \frac{3}{e}$. So $\frac{2}{e} < S < \frac{3}{e}$, roughly between $0.74$ and $1.10$ with $e \approx 2.72$. (The exact sum is $\frac{e}{(e - 1)^2}$, about $0.92$, but that is another chapter.)

$f(x) = \frac{e^{1/x}}{x^2}$ is continuous and positive on $[1, \infty)$. For the decrease no derivative is needed: $x \mapsto \frac{1}{x}$ decreases and $e^u$ increases, so $e^{1/x}$ DECREASES; $\frac{1}{x^2}$ decreases too; and a product of two POSITIVE decreasing functions is decreasing. Saying so in one sentence is a complete justification. With $u = \frac{1}{x}$, $du = -\frac{dx}{x^2}$, the bounds $x = 1$ and $x = t$ become $u = 1$ and $u = \frac{1}{t}$: $\int_1^t \frac{e^{1/x}}{x^2}\,dx = \int_{1/t}^{1} e^u\,du = e - e^{1/t} \to e - 1$. The integral converges, so the series CONVERGES by the integral test. The product rule for monotonicity needs POSITIVE factors: $(-x)(-x) = x^2$ is a product of two decreasing functions that increases on $[1, \infty)$.

For $p > 0$, let $f(x) = \frac{1}{x^p} = x^{-p}$. It is continuous and positive on $[1, \infty)$, and $f'(x) = -p x^{-p-1} < 0$: decreasing. The integral test applies, so $\sum \frac{1}{n^p}$ behaves like $\int_1^{\infty} \frac{dx}{x^p}$: convergent for $p > 1$, divergent for $0 < p \le 1$. For $p \le 0$ the hypothesis DECREASING fails: $f(x) = x^{-p} = x^{|p|}$ is constant ($p = 0$) or increasing ($p < 0$), and a test whose hypothesis fails says nothing. The right tool is the divergence test: $\frac{1}{n^p} = n^{|p|}$ tends to $\infty$ or equals $1$, never to $0$, so the series diverges. Together: $\sum \frac{1}{n^p}$ converges exactly when $p > 1$.

$\frac{1}{\sqrt[3]{n^2}} = \frac{1}{n^{2/3}}$: $p = \frac{2}{3} \le 1$, DIVERGES. $\frac{n\sqrt{n}}{n^3} = \frac{n^{3/2}}{n^3} = \frac{1}{n^{3/2}}$: $p = \frac{3}{2} > 1$, CONVERGES. $\frac{\sqrt[4]{n}}{n} = \frac{n^{1/4}}{n} = \frac{1}{n^{3/4}}$: $p = \frac{3}{4}$, DIVERGES. $n^{-\pi/3} = \frac{1}{n^{\pi/3}}$ with $\frac{\pi}{3} > 1$ because $\pi > 3$: CONVERGES, with no decimal needed. The exponent is read AFTER the algebra: $\sqrt[3]{n^2}$ is $n^{2/3}$, not $n^{3/2}$, and that single swap turns a divergent series into a convergent one.

$3^{-\ln n} = e^{-\ln 3 \cdot \ln n} = \left(e^{\ln n}\right)^{-\ln 3} = n^{-\ln 3}$. So $\sum 3^{-\ln n}$ is a $p$-series with $p = \ln 3$. Since $3 > e$ and $\ln$ increases, $\ln 3 > \ln e = 1$: it CONVERGES. In the same way $2^{-\ln n} = n^{-\ln 2}$, and $2 < e$ gives $\ln 2 < 1$: $\sum 2^{-\ln n}$ DIVERGES. The dividing base is $e$ itself, for which $e^{-\ln n} = \frac{1}{n}$, the harmonic series. A term like $3^{-\ln n}$ LOOKS geometric, but the variable is inside a logarithm, not alone in the exponent, and it decreases far more slowly than $\left(\frac{1}{3}\right)^n$.

With $k = n + 3$: $\sum_{n=1}^{\infty} \frac{1}{(n + 3)^{3/2}} = \sum_{k=4}^{\infty} \frac{1}{k^{3/2}}$, a $p$-series with $p = \frac{3}{2}$ missing its first three terms: it CONVERGES, since dropping finitely many terms never changes convergence. For $\sum \frac{1}{2n - 1}$: $f(x) = \frac{1}{2x - 1}$ is continuous, positive and decreasing on $[1, \infty)$, and $\int_1^t \frac{dx}{2x - 1} = \frac{1}{2}\ln(2t - 1) \to \infty$: the series DIVERGES. The classmate's verdict is right, his words are not: the odd reciprocals are not half of the harmonic series term by term, and half of infinity is not a number to converge to. What is true is $\frac{1}{2n - 1} \ge \frac{1}{2n}$, a comparison with a divergent series, which is the next exercise.

$\sum_{n=1}^{\infty} \frac{1}{n^2} \ge 1 + \frac{1}{4} > 1$, since all the terms are positive: the sum is not $1$. The integral is an AREA under a curve, the series is a sum of rectangles, and Exercise 1 d) compares them in the right direction: rectangles of heights $a_2, a_3, \dots$ placed on $[1, 2], [2, 3], \dots$ fit under the curve, so $\sum_{n=2}^{\infty} \frac{1}{n^2} \le \int_1^{\infty} \frac{dx}{x^2} = 1$, and $\sum_{n=1}^{\infty} \frac{1}{n^2} \le 1 + 1 = 2$. So $\frac{5}{4} \le S \le 2$. (Euler proved that $S = \frac{\pi^2}{6}$, about $1.64$.) Exercise 5 narrows this interval to a width of $\frac{1}{12}$.

Positivity first: $\cos n \ge -1$, so $1 + \cos n \ge 0$ and every term is $\ge 0$. Then $1 + \cos n \le 2$ gives $0 \le \frac{1 + \cos n}{n^{3/2}} \le \frac{2}{n^{3/2}}$ for all $n \ge 1$. The series $\sum \frac{2}{n^{3/2}}$ is twice a $p$-series with $p = \frac{3}{2} > 1$: it converges. The terms are SMALLER than those of a convergent series, so $\sum \frac{1 + \cos n}{n\sqrt{n}}$ CONVERGES by the comparison test. The integral test is out: $f(x) = \frac{1 + \cos x}{x^{3/2}}$ is not decreasing, since $f(\pi) = 0$ while $f(2\pi) = \frac{2}{(2\pi)^{3/2}} > 0$. A bounded oscillating factor is the typical sign that the comparison test, not the integral test, is the tool.

Her inequality is true, and it proves nothing: being SMALLER than the terms of a divergent series is compatible with both verdicts, since $\frac{1}{n^2} < \frac{1}{n}$ too and $\sum \frac{1}{n^2}$ converges. To prove divergence she needs a LOWER bound by a divergent series. On the figure the orange terms sit ABOVE the green curve $\frac{1}{3x}$: indeed $\sqrt{n} \le n$ for $n \ge 1$, so $2n + \sqrt{n} \le 3n$ and $\frac{1}{2n + \sqrt{n}} \ge \frac{1}{3n}$. Since $\sum \frac{1}{3n} = \frac{1}{3}\sum \frac{1}{n}$ diverges (harmonic), $\sum \frac{1}{2n + \sqrt{n}}$ DIVERGES by the comparison test. The method: to show divergence, make the denominator BIGGER, never smaller; the constant $\frac{1}{3}$ costs nothing.

For $n \ge 3$, $n^2 - 5 > 0$ and the terms are positive. But $n^2 - 5 < n^2$ gives $\frac{1}{n^2 - 5} > \frac{1}{n^2}$: bigger than a CONVERGENT series, no conclusion. The repair is to give up a constant: $n^2 - 5 \ge \frac{n^2}{2}$ is equivalent to $\frac{n^2}{2} \ge 5$, that is $n^2 \ge 10$, true for every $n \ge 4$. Then $\frac{1}{n^2 - 5} \le \frac{2}{n^2}$ for $n \ge 4$, and $\sum \frac{2}{n^2}$ converges ($p = 2$), so the series CONVERGES. At $n = 3$: $n^2 - 5 = 4 < \frac{9}{2}$, the inequality fails, and it does not matter, since the test only asks for $n \ge N$ and one term is a finite number. Exercise 4 reaches the same verdict in one limit.

All terms are positive. Making the denominator smaller makes the fraction bigger: $3^n + n > 3^n$, so $\frac{2^n}{3^n + n} < \frac{2^n}{3^n} = \left(\frac{2}{3}\right)^n$. The geometric series $\sum \left(\frac{2}{3}\right)^n$ has ratio $r = \frac{2}{3}$, with $|r| < 1$: it converges. The terms are smaller than those of a convergent series, so the series CONVERGES. Here the easy inequality is also the right one: dropping a positive term from the denominator gives an UPPER bound, exactly what convergence needs.

For $n \ge 3$: $3 > e$, so $\ln n \ge \ln 3 > \ln e = 1$, and $\frac{\ln n}{\sqrt{n}} \ge \frac{1}{\sqrt{n}}$. The series $\sum \frac{1}{\sqrt{n}}$ is a $p$-series with $p = \frac{1}{2} \le 1$: it diverges. The terms are bigger than those of a divergent series from $n = 3$ on, so $\sum \frac{\ln n}{\sqrt{n}}$ DIVERGES. The range matters and must be written: at $n = 1$, $\ln 1 = 0$, and at $n = 2$, $\ln 2 < 1$, so the inequality is false there. Writing it for all $n \ge 1$ puts a false statement on the page, and markers take the mark for it even when the verdict is right.

The terms are positive. Dominant terms: $2n^2$ on top, $\sqrt{n^7} = n^{7/2}$ below, so $a_n$ behaves like $\frac{2n^2}{n^{7/2}} = \frac{2}{n^{3/2}}$. Take $b_n = \frac{1}{n^{3/2}}$ (the constant $2$ can stay out of $b_n$, the limit collects it). Writing $\sqrt{n^7 + 1} = n^{7/2}\sqrt{1 + \frac{1}{n^7}}$: $\frac{a_n}{b_n} = \frac{(2n^2 + 3n)\,n^{3/2}}{n^{7/2}\sqrt{1 + 1/n^7}} = \frac{2 + \frac{3}{n}}{\sqrt{1 + \frac{1}{n^7}}} \to 2$. Since $0 < 2 < \infty$ and $\sum \frac{1}{n^{3/2}}$ converges ($p = \frac{3}{2} > 1$), the series CONVERGES by the limit comparison test. The algebra is always the same: factor the dominant power out of every sum BEFORE taking the limit.

$n + 5 > 0$, and $n^2 - 3n + 7$ has discriminant $9 - 28 = -19 < 0$ with a positive leading coefficient, so it is positive for every $n$: all terms are positive, which the test requires. Dominant terms: $\frac{n}{n^2} = \frac{1}{n}$. With $b_n = \frac{1}{n}$: $\frac{a_n}{b_n} = \frac{n^2 + 5n}{n^2 - 3n + 7} = \frac{1 + \frac{5}{n}}{1 - \frac{3}{n} + \frac{7}{n^2}} \to 1$. The harmonic series diverges, so the series DIVERGES. A direct comparison would need an inequality such as $n^2 - 3n + 7 \le 2n^2$, true but one more thing to prove: that is the time the limit version saves.

The terms are positive. In the numerator, $\sqrt{n}$ dominates $\ln n$: by L'Hospital's rule, $\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} = \lim \frac{1/x}{1/(2\sqrt{x})} = \lim \frac{2}{\sqrt{x}} = 0$, and the limit of the sequence is the limit of the function. So $a_n$ behaves like $\frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}$. With $b_n = \frac{1}{n^{3/2}}$: $\frac{a_n}{b_n} = \frac{(\sqrt{n} + \ln n)\,n^{3/2}}{n^2 + 1} = \frac{1 + \frac{\ln n}{\sqrt{n}}}{1 + \frac{1}{n^2}} \to 1$. Since $\sum \frac{1}{n^{3/2}}$ converges, the series CONVERGES. A logarithm is never the dominant term against a positive power of $n$, and it has to be SAID, with the limit, not assumed.

With $b_n = \frac{1}{n^2}$: $\frac{a_n}{b_n} = \ln n \to \infty$, and $\sum \frac{1}{n^2}$ CONVERGES. A term infinitely larger than that of a convergent series can do anything: no conclusion. With $b_n = \frac{1}{n}$: $\frac{a_n}{b_n} = \frac{\ln n}{n} \to 0$, and $\sum \frac{1}{n}$ DIVERGES. A term infinitely smaller than that of a divergent series can do anything: no conclusion. The logarithm sits between the two powers, so the benchmark must sit between them too: $b_n = \frac{1}{n^{3/2}}$ gives $\frac{a_n}{b_n} = \frac{\ln n}{n^{1/2}} \to 0$ (part c), and $\sum \frac{1}{n^{3/2}}$ converges, so the one-sided case applies: $\sum \frac{\ln n}{n^2}$ CONVERGES. Any exponent strictly between $1$ and $2$ works; $\frac{3}{2}$ is the one that keeps the arithmetic clean.

It is not a $p$-series: in $\frac{1}{n^p}$ the exponent is a CONSTANT, and here it changes with $n$ and tends to $1$. With $b_n = \frac{1}{n}$: $\frac{a_n}{b_n} = \frac{n}{n^{1 + 1/n}} = n^{-1/n} = \frac{1}{n^{1/n}}$. Now $\ln\left(n^{1/n}\right) = \frac{\ln n}{n} \to 0$, so $n^{1/n} \to e^0 = 1$ and the ratio tends to $1$. Since $0 < 1 < \infty$ and the harmonic series diverges, $\sum \frac{1}{n^{1 + 1/n}}$ DIVERGES. The extra factor $n^{-1/n}$ tends to $1$: it is far too weak to pull the harmonic series into convergence, although every single exponent exceeds $1$.