Corrected exercises: the ratio and root tests, and choosing a test (MATH 141)

Terms positive, so no absolute values are needed. $\frac{a_{n+1}}{a_n} = \frac{(n+1)^3}{3^{n+1}} \cdot \frac{3^n}{n^3} = \frac{1}{3}\left(\frac{n+1}{n}\right)^3 = \frac{1}{3}\left(1 + \frac{1}{n}\right)^3$. As $n \to \infty$, $1 + \frac{1}{n} \to 1$, so $L = \frac{1}{3}$. Since $L < 1$, the series converges by the ratio test. The gesture that earns the mark is the rewriting $\frac{3^n}{3^{n+1}} = \frac{1}{3}$ and $\frac{(n+1)^3}{n^3} = \left(1 + \frac{1}{n}\right)^3$ BEFORE the limit: the powers of $n$ contribute a factor that tends to $1$, the exponential contributes the whole of $L$. That is the general pattern: a polynomial factor never changes the value of $L$.

$\frac{a_{n+1}}{a_n} = \frac{100^{n+1}}{(n+1)!} \cdot \frac{n!}{100^n} = \frac{100}{n+1}$, using $(n+1)! = (n+1)\cdot n!$. So $L = \lim_{n\to\infty}\frac{100}{n+1} = 0 < 1$: the series converges. The ratio is $\ge 1$ exactly when $n + 1 \le 100$, so the terms increase up to $a_{99} = a_{100}$ and only then decrease, and the terms around $n = 100$ are astronomically large. None of this matters: convergence is a property of the TAIL of the series, and removing or changing finitely many terms changes the sum but never whether it exists. The factorial grows faster than any geometric sequence, however large its base, and the ratio test shows it in one line.

$\frac{a_{n+1}}{a_n} = \frac{((n+1)!)^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2} = \frac{(n+1)^2}{(2n+2)(2n+1)}$, because $\frac{(n+1)!}{n!} = n+1$ (squared) and $(2n+2)! = (2n+2)(2n+1)(2n)!$. Simplify: $\frac{(n+1)^2}{2(n+1)(2n+1)} = \frac{n+1}{2(2n+1)} \to \frac{1}{4}$ (divide top and bottom by $n$: $\frac{1 + 1/n}{4 + 2/n}$). $L = \frac{1}{4} < 1$: converges. The classic slip is $\frac{(2n+2)!}{(2n)!} = 2n+2$: going from $(2n)!$ to $(2n+2)!$ adds TWO factors, and forgetting $(2n+1)$ gives a limit of $\infty$ and the opposite verdict.

$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{(n+1)\,n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \frac{1}{\left(1 + \frac{1}{n}\right)^n}$. The limit $\left(1 + \frac{1}{n}\right)^n \to e$ is the one of the sequences chapter, so $L = \frac{1}{e}$, about $\frac{1}{2.72} \approx 0.37 < 1$: converges. The trap here is the form $1^\infty$: $\frac{n}{n+1} \to 1$, and writing $\left(\frac{n}{n+1}\right)^n \to 1^\infty = 1$ would give $L = 1$ and a false no information. The exponent grows with $n$, so the base tending to $1$ decides nothing; the limit is $e^{-1}$.

The terms are $a_1 = \frac{1}{3}$, $a_2 = \frac{8}{9}$, $a_3 = 1$, then $a_4 = \frac{64}{81}$ and they decrease: the ratios $\frac{8}{3}$ and $\frac{9}{8}$ are above $1$, the third, $\frac{64}{81}$, is below. There is no contradiction: the ratio test reads the LIMIT of the ratios, which is $\frac{1}{3}$, and the proof only uses the ratios from some rank $N$ on. On the figure, from $n = 3$ on every ratio is below $1$ and they settle on the dashed line at $\frac{1}{3}$. A student who stops at $\frac{a_2}{a_1} = \frac{8}{3} > 1$ and concludes divergence has read one ratio instead of a limit, which the test never allows.

The terms are positive. $\sqrt[n]{a_n} = \frac{2n+3}{3n+1} = \frac{2 + 3/n}{3 + 1/n} \to \frac{2}{3}$. $L = \frac{2}{3} < 1$: the series converges by the root test. With the ratio test the same series would require the quotient of $\left(\frac{2n+5}{3n+4}\right)^{n+1}$ by $\left(\frac{2n+3}{3n+1}\right)^n$, a $1^\infty$ form that is painful to settle. The form of the term, a fraction raised to the power $n$, chose the test.

$\sqrt[n]{a_n} = \frac{1}{3}\left(1 + \frac{1}{n}\right)^{n^2/n} = \frac{1}{3}\left(1 + \frac{1}{n}\right)^n \to \frac{e}{3}$. Since $e < 3$, $L = \frac{e}{3} < 1$, about $0.91$: converges. Two traps. The exponent: $\left(b^{n^2}\right)^{1/n} = b^{n}$, not $b$; the $n$-th root divides the exponent by $n$, it does not remove it. And the limit: $\left(1 + \frac{1}{n}\right)^n$ is a $1^\infty$ form whose limit is $e$, not $1$. Note that $\left(1 + \frac{1}{n}\right)^{n^2}$ alone tends to infinity: the series converges only because $3^n$ grows faster, and $e < 3$ is exactly the comparison that decides.

$\sqrt[n]{a_n} = \frac{2}{\left(n^{1/n}\right)^{10}} \to \frac{2}{1^{10}} = 2$, since $n^{1/n} \to 1$. $L = 2 > 1$: DIVERGES. The terms start at $a_1 = 2$, fall below $1$ from $n = 2$ ($a_2 = \frac{4}{1024}$) and stay below $1$ until $n = 58$, because $2^n < n^{10}$ there; from $n = 59$ on, $2^n > n^{10}$ and the terms grow without bound. A table of the first fifty terms would suggest convergence, and it proves nothing: no finite number of terms decides a series. The power $n^{10}$, however large its exponent, contributes $\left(n^{1/n}\right)^{10} \to 1$ to $L$.

$\sqrt[n]{a_n} = \frac{\arctan n}{2}$, and $\arctan n \to \frac{\pi}{2}$ as $n \to \infty$, so $L = \frac{\pi}{4}$. Since $\pi < 4$, $L < 1$ (about $0.79$): converges. The limit $\frac{\pi}{2}$ is exact; no decimal of $\arctan$ is needed. Note that each root $\frac{\arctan n}{2}$ is below $\frac{\pi}{4}$, and the test does not care: only the limit is compared with $1$.

$a_n < \left(\frac{4}{5}\right)^n$ is equivalent, taking $n$-th roots of positive numbers, to $\frac{2n+3}{3n+1} < \frac{4}{5}$, that is $5(2n+3) < 4(3n+1)$, $10n + 15 < 12n + 4$, $n > \frac{11}{2}$. So $N = 6$: the inequality holds for all $n \ge 6$ and fails at $n = 5$, where $\frac{13}{16} > \frac{4}{5}$ (on the figure, the points pass under the curve at $n = 6$). Then $0 < a_n < \left(\frac{4}{5}\right)^n$ for $n \ge 6$, and $\sum_{n=6}^{\infty}\left(\frac{4}{5}\right)^n = \frac{(4/5)^6}{1 - 4/5} = 5\left(\frac{4}{5}\right)^6 = 4\left(\frac{4}{5}\right)^5$ is a convergent geometric series; by comparison the tail converges, and adding the five first terms does not change that. This IS the proof of the root test: any $r$ strictly between $L$ and $1$ works, here $r = \frac{4}{5}$ between $\frac{2}{3}$ and $1$. It also explains why $L = 1$ gives nothing: then no fixed $r < 1$ can be squeezed in.

For $a_n = \frac{1}{n}$: $\frac{a_{n+1}}{a_n} = \frac{n}{n+1} \to 1$ and $\sqrt[n]{a_n} = \frac{1}{n^{1/n}} \to 1$. For $a_n = \frac{1}{n^2}$: $\frac{a_{n+1}}{a_n} = \left(\frac{n}{n+1}\right)^2 \to 1$ and $\sqrt[n]{a_n} = \frac{1}{\left(n^{1/n}\right)^2} \to 1$. Four limits equal to $1$, so both tests are inconclusive for both series. The verdicts come from the $p$-series test: $\sum\frac{1}{n}$ diverges ($p = 1$) and $\sum\frac{1}{n^2}$ converges ($p = 2 > 1$). On the figure the two families of ratios are indistinguishable in the limit, and yet the series behave oppositely: this is the proof that $L = 1$ cannot carry a verdict.

The ratio test needs the LIMIT to be below $1$, not each ratio. The proof of the test works with a fixed $r$ such that $\frac{a_{n+1}}{a_n} \le r < 1$ for all large $n$, which gives $a_n \le a_N r^{n-N}$ and a geometric comparison. Here the ratios $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \dots$ creep up to $1$ and eventually exceed every fixed $r < 1$: the terms do decrease, but more and more slowly, not geometrically, and the harmonic series diverges. Writing the ratio is less than 1, so it converges is the most frequent wrong line on this chapter and it scores zero.

Say $P$ has degree $p$ and leading coefficient $b$, $Q$ has degree $q$ and leading coefficient $c$. Then $\frac{a_{n+1}}{a_n} = \frac{P(n+1)}{P(n)} \cdot \frac{Q(n)}{Q(n+1)}$. Dividing top and bottom by $n^p$, $\frac{P(n+1)}{P(n)} \to \frac{b}{b} = 1$, because $(n+1)^p$ and $n^p$ have the same leading behaviour; likewise $\frac{Q(n)}{Q(n+1)} \to 1$. So $L = 1$ for EVERY rational term, whatever the degrees: the ratio test is always wasted on it, and so is the root test. The right tool is a limit comparison with $\frac{1}{n^{q-p}}$, the ratio of the leading terms, and then the $p$-series test: convergence exactly when $q - p > 1$. Recognising the form saves the whole computation.

The term is algebraic in $n$, so by c) the ratio test would give $1$. Dominant terms: $\frac{\sqrt{n}}{n^2 + 1}$ behaves like $\frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}}$. Limit comparison with $b_n = \frac{1}{n^{3/2}}$, both positive: $\frac{a_n}{b_n} = \frac{n^{1/2} \cdot n^{3/2}}{n^2 + 1} = \frac{n^2}{n^2 + 1} \to 1$, a finite positive number. Since $\sum\frac{1}{n^{3/2}}$ converges ($p = \frac{3}{2} > 1$), the series converges. The hypotheses checked on the copy: positive terms, a limit that is finite AND nonzero.

$\sqrt[n]{a_n} = \frac{n+1}{n+2} \to 1$: the root test is inconclusive. Every root is below $1$, which again means nothing. Go back to the first question of any strategy: do the terms tend to $0$? Write $\frac{n+1}{n+2} = 1 - \frac{1}{n+2}$, so $a_n = \left(1 - \frac{1}{n+2}\right)^{n+2} \cdot \left(1 - \frac{1}{n+2}\right)^{-2}$. The first factor tends to $e^{-1}$ (the limit $\left(1 + \frac{x}{m}\right)^m \to e^x$ with $x = -1$, $m = n + 2$) and the second to $1$. So $a_n \to \frac{1}{e} \ne 0$ and the series DIVERGES by the divergence test. A term to the power $n$ invited the root test, and the root test was silent; the $1^\infty$ form of the term itself was the information.

$\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{3^{n+1}}{(n+1)4^{n+1}} \cdot \frac{n\,4^n}{3^n} = \frac{3}{4} \cdot \frac{n}{n+1} \to \frac{3}{4}$. $L = \frac{3}{4} < 1$: the series converges ABSOLUTELY, hence converges. The sign $(-1)^n$ disappears in the absolute value; there is no need to check that the terms decrease or to invoke the alternating series test, and doing so costs time without earning a mark. The correct conclusion names absolute convergence, which is stronger than convergence.

$\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{4^{n+1}}{(n+1)^4} \cdot \frac{n^4}{4^n} = 4\left(\frac{n}{n+1}\right)^4 \to 4$. $L = 4 > 1$: the series DIVERGES. On the figure, $\lvert a_1\rvert = 4$, $\lvert a_2\rvert = 1$, $\lvert a_3\rvert = \frac{64}{81}$, $\lvert a_4\rvert = 1$, then $\frac{1024}{625}$, $\frac{256}{81}$, and the terms keep growing. The ratio $4\left(\frac{n}{n+1}\right)^4$ is above $1$ as soon as $\sqrt{2}\,n > n + 1$ (take fourth roots: $4^{1/4} = \sqrt{2}$), that is $n > \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$, so from $n = 3$ on each term is larger in absolute value than the previous one. The terms, far from tending to $0$, tend to infinity in absolute value, while the signs alternate: it is not an alternating series question at all.

$\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \left(\frac{n}{n+1}\right)^{1/3} \to 1$ and $\sqrt[n]{\lvert a_n\rvert} = n^{-1/(3n)} \to 1$: both tests are inconclusive. Alternating series test with $b_n = \frac{1}{\sqrt[3]{n}}$: $b_n > 0$, $b_{n+1} < b_n$ because the cube root is increasing, and $b_n \to 0$. So $\sum\frac{(-1)^n}{\sqrt[3]{n}}$ converges. Absolute series: $\sum\frac{1}{n^{1/3}}$ is a $p$-series with $p = \frac{1}{3} \le 1$, it diverges. The series is CONDITIONALLY convergent. The ratio test, silent here, can never detect conditional convergence: it only ever sees $\lvert a_n\rvert$.

Choose $r$ with $1 < r < L$ (for instance $r = \frac{1 + L}{2}$; if $L = \infty$ take $r = 2$). Since the ratios tend to $L$, there is $N$ such that $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert > r > 1$ for all $n \ge N$. Then $\lvert a_{n+1}\rvert > \lvert a_n\rvert$ for $n \ge N$, so $\lvert a_n\rvert \ge \lvert a_N\rvert > 0$ for all $n \ge N$ (in fact $\lvert a_n\rvert \ge \lvert a_N\rvert r^{n-N} \to \infty$). The terms do not tend to $0$, and $\sum a_n$ diverges by the divergence test, whatever the signs. Conditional convergence requires $\sum a_n$ to converge, hence $a_n \to 0$: impossible here. So after a ratio or root limit above $1$, the conclusion is simply diverges, never let us try the alternating series test.

Write $a_n = (-1)^n\frac{n!}{1 \cdot 3 \cdots (2n-1)}$. Going from $n$ to $n + 1$ multiplies the numerator by $n + 1$ and the denominator by the next odd number, $2n + 1$: $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{n+1}{2n+1} \to \frac{1}{2}$. $L = \frac{1}{2} < 1$: absolutely convergent. Terms defined by a product with a growing number of factors are the ratio test's home ground: the product collapses to its last factor. Writing both products out to their last factors, $1 \cdot 3 \cdots (2n-1)(2n+1)$ over $1 \cdot 3 \cdots (2n-1)$, is what makes the cancellation visible. Confusing the odd product with $(2n)!$ adds a spurious factor $2n + 2$ and gives $L = 0$: the right verdict for a wrong reason, which loses the method mark.

No formula for $a_n$ is needed: the recursion gives the ratio directly. $a_1 > 0$ and the factor $\frac{5n+1}{4n+3}$ is positive, so all terms are positive, and $\frac{a_{n+1}}{a_n} = \frac{5n+1}{4n+3} \to \frac{5}{4} > 1$: the series DIVERGES by the ratio test. As in Exercise 4 d), the terms themselves confirm it: the factor is $\ge 1$ as soon as $5n + 1 \ge 4n + 3$, i.e. $n \ge 2$, so $a_2 = \frac{12}{7}$, $a_3 = \frac{11}{11}a_2 = \frac{12}{7}$, and the terms never go below $\frac{12}{7}$ afterwards. A recursive definition is the clearest signal of the ratio test on a paper: the quotient is written in the statement.

Since $-1 \le \cos n \le 1$, the factor satisfies $0 < \frac{1}{\sqrt{n}} \le \frac{2 + \cos n}{\sqrt{n}} \le \frac{3}{\sqrt{n}}$, so all terms are positive. The ratio $\frac{a_{n+1}}{a_n} = \frac{2 + \cos n}{\sqrt{n}}$ has no simple formula for its limit, because $\cos n$ itself has no limit, but it is squeezed: $0 < \frac{a_{n+1}}{a_n} \le \frac{3}{\sqrt{n}} \to 0$. By the squeeze theorem $L = 0 < 1$ and the series converges. The oscillating numerator is harmless because it is BOUNDED and divided by something that goes to infinity; the squeeze is the sentence that makes the limit legitimate.

The terms are $a_n = \frac{1}{3^n}$ for odd $n$ and $a_n = \frac{3}{3^n}$ for even $n$. For odd $n$: $\frac{a_{n+1}}{a_n} = \frac{3/3^{n+1}}{1/3^n} = 1$; for even $n$: $\frac{a_{n+1}}{a_n} = \frac{1/3^{n+1}}{3/3^n} = \frac{1}{9}$. The ratios alternate $1, \frac{1}{9}, 1, \frac{1}{9}, \dots$: they have no limit, so the ratio test does not apply (and certainly does not say diverges because some ratios equal $1$). Root test: $\sqrt[n]{a_n} = \frac{(2 + (-1)^n)^{1/n}}{3}$, and $1 \le (2 + (-1)^n)^{1/n} \le 3^{1/n} \to 1$, so by the squeeze theorem $L = \frac{1}{3} < 1$: converges. The sum: $\sum_{n=1}^{\infty}\frac{2}{3^n} + \sum_{n=1}^{\infty}\left(-\frac{1}{3}\right)^n = 2 \cdot \frac{1/3}{1 - 1/3} + \frac{-1/3}{1 + 1/3} = 1 - \frac{1}{4} = \frac{3}{4}$, splitting being legitimate because both geometric series converge. The root test is the stronger of the two: whenever the ratio limit exists the root limit exists and is equal, but not conversely.