Choosing a convergence test
Every convergence test answers the same question, but each one answers it only for series of a certain shape. Looking at a series, how do you tell which test will settle it, and what do you do when the first test you try says nothing?
Where you are 6 in Convergence tests
When you reach for it
By the end of this unit there are nine ways to decide whether \sum a_n converges: the The nth term test for divergence, the geometric series test, the integral test and the p series test, the direct and limit comparison tests, the alternating series test, and the ratio and root tests. Each one was taught on series built for it. A series met in the wild comes with no label, and a test applied to the wrong shape usually does not give a wrong answer: it gives no answer, after a page of algebra.
So the skill is reading. The form of a_n (what grows, what shrinks, whether the sign changes, whether the whole term is raised to a power) points at the test whose hypotheses it meets and whose conclusion it can reach. This lesson is about that reading. Day 1 picks a test from the shape of the term and compares routes when more than one works. Day 2 handles series on which the most tempting test is silent, and sorts alternating series into absolutely and conditionally convergent ones.
The procedure
Day 1: reading the shape of the term
Look at the size of the terms first. If a_n does not tend to 0, the series diverges, and nothing else is needed. This costs one limit, often one glance: a term that is at least 1 for every n, or a ratio of polynomials of equal degree, settles the question at once. The test is one-sided. A limit of 0 says nothing, and the reading goes on.
Recognize the two families with a verdict by form. A geometric series \sum c\,r^n converges exactly when \lvert r \rvert < 1, and a p series \displaystyle \sum \frac{1}{n^p} exactly when p > 1. Both need something constant: the ratio r, the exponent p. Rewriting the term with the laws of exponents often reveals one of them, and then no test is run at all.
Find the part that controls the size. For large n, quantities grow at sharply different rates (limits of the kind found in Convergence of sequences): for constants q > 0, p > 0 and b > 1,
where \ll means the ratio of the left side to the right side tends to 0. In a sum, the fastest-growing piece dominates; in a product, every factor counts. Keeping only the dominant pieces of the numerator and denominator gives a simpler term, and the simpler term tells you which test to use:
- If the simpler term is a p series or a geometric series, compare with it. Limit comparison needs only the limit of a quotient; direct comparison needs an inequality, which is quicker when one is obvious (a bounded factor such as \sin n or \arctan n).
- If the term carries a factorial or an exponential c^n, the ratio test is natural: factorials cancel in \displaystyle \frac{a_{n+1}}{a_n} and c^{n+1}/c^n = c.
- If the whole term is something raised to the nth power, or to a power such as n^2, the root test undoes that power.
- If the term is f(n) for a positive, decreasing f with an antiderivative you can find (often a function and its derivative, as in \displaystyle \frac{1}{x\ln x}), the integral test applies.
- If the signs change, test \sum \lvert a_n \rvert first with the tests above; the alternating series test comes in only if that fails.
The table collects what each test needs, what it concludes, and where it is silent. Silence is the reason to know more than one test.
| Test | Needs | Concludes, and where it is silent |
|---|---|---|
| nth term | nothing | diverges if a_n \not\to 0; silent if a_n \to 0 |
| Geometric | a fixed ratio r | converges if and only if \lvert r \rvert < 1 |
| p series | a fixed exponent p | converges if and only if p > 1 |
| Integral | f positive, continuous, eventually decreasing | same verdict as \int_1^{\infty} f(x)\,dx |
| Direct comparison | 0 \le a_n \le b_n, or a_n \ge b_n \ge 0 | below a convergent series: converges; above a divergent one: diverges; silent otherwise |
| Limit comparison | a_n > 0, b_n > 0, \displaystyle \frac{a_n}{b_n} \to c | 0 < c < \infty: same verdict as \sum b_n; silent if c = 0 and \sum b_n diverges, or c = \infty and \sum b_n converges |
| Alternating series | b_n eventually decreasing to 0 | converges; never shows divergence or absolute convergence |
| Ratio | a_n \ne 0 | L < 1: converges absolutely; L > 1: diverges; silent if L = 1 or no limit |
| Root | nothing | the same as the ratio test |
Two tests may both apply. Then the choice is about length: which route needs fewer side facts (a derivative to show decrease, an inequality to prove, a hard limit), and which route gives more than a verdict, such as the sum or a bound on it.
Day 2: when the first choice is silent
A silent test is information. The ratio test gives L = 1 on every rational term (The ratio and root tests, Exercise 11 there), and the same happens for terms built from powers of n and logarithms, because such terms shrink more slowly than any geometric sequence. That limit says the series must be compared with a p series, or tested by integration, not by a geometric yardstick.
Comparison can fail in two ways. The inequality can point the wrong way: a term smaller than a divergent series, or larger than a convergent one, proves nothing. And the limit in limit comparison can be 0 or \infty on the side that decides nothing. Both failures mean the benchmark was too far from the term, and they also say in which direction to move it: a term that is smaller than \displaystyle \frac1n but larger than every \displaystyle \frac{1}{n^p} with p > 1 lies in the narrow band between, where no p series can serve as a benchmark and a sharper tool, such as the integral test, is needed (Example 6).
For a series with both signs, the order matters. First test \sum \lvert a_n \rvert, a series of positive terms, with the Day 1 reading. If it converges, \sum a_n converges absolutely and the work is done. If \lvert a_n \rvert \not\to 0, the series diverges by the nth term test. Only when \sum \lvert a_n \rvert diverges while \lvert a_n \rvert \to 0 is the alternating series test needed, and if it applies, the convergence is conditional.
Worked examples
Day 1: three series settled by their form
Decide whether each series converges: (a) \displaystyle \sum_{n=1}^{\infty} \sqrt[n]{n}; (b) \sum_{n=1}^{\infty} \dfrac{2^{n+1}}{5^{n-1}}; (c) \displaystyle \sum_{n=1}^{\infty} \dfrac{\sqrt[3]{n}}{n^2}.
Solution(a) Look at the size of the terms first. For n \ge 1 the number n is at least 1, so its nth root is at least 1. Every term is at least 1, the terms do not tend to 0, and the series diverges by the nth term test; indeed S_n \ge n. No limit had to be computed. (The terms do tend to 1, since \displaystyle \ln \sqrt[n]{n} = \frac{\ln n}{n} \to 0, but the bound is enough.)
(b) Both powers have exponents linear in n, which suggests a geometric series. By the laws of exponents,
so the series is geometric with first term 4 and ratio \displaystyle \frac25. Since \displaystyle \left\lvert \frac25 \right\rvert < 1, it converges, with sum \dfrac{4}{1 - 2/5} = \dfrac{20}{3}.
(c) Write both powers of n as powers: \displaystyle \frac{n^{1/3}}{n^2} = \frac{1}{n^{5/3}}. This is a p series with \displaystyle p = \frac53 > 1, so it converges.
The ratio test would have given L = 1 on (c), which decides nothing.
Day 1: an exponent that drifts toward 1
Decide whether \sum_{n=2}^{\infty} \dfrac{1}{n^{1+1/\ln n}} converges.
SolutionThe exponent \displaystyle p_n = 1 + \frac{1}{\ln n} is greater than 1 for every n \ge 2, which makes the series look like a convergent p series. But the p series test needs one fixed exponent, and here the exponent changes with n and falls toward 1 (Figure 1). The test does not apply.
Split off the part that drifts: n^{1+1/\ln n} = n \cdot n^{1/\ln n}. The second factor is a constant in disguise. Since n = e^{\ln n},
So a_n = \dfrac{1}{e\,n} for every n \ge 2, the harmonic term times the constant \displaystyle \frac1e. Since \displaystyle \sum_{n=2}^{\infty} \frac1n diverges, so does \displaystyle \frac1e times it, and the series diverges. Limit comparison with \displaystyle \frac1n says the same, with the quotient equal to \displaystyle \frac1e for every n.
The exponents are p_2 \approx 2.443, p_{10} \approx 1.434 and p_{60} \approx 1.244, all above 1. A fixed exponent of 1.244 would give a convergent series; an exponent that keeps falling toward 1 does not, because its excess \displaystyle \frac{1}{\ln n} shrinks so fast that n^{1/\ln n} never grows past e.
Day 1: Malakai’s two routes
Malakai is asked whether \sum_{n=1}^{\infty} \dfrac{5^n + n}{n!} converges. Settle it with the ratio test, then by splitting the term, and say which route is shorter.
SolutionThe factorial points to the ratio test. With \displaystyle a_n = \frac{5^n + n}{n!} and \displaystyle \frac{n!}{(n+1)!} = \frac{1}{n+1},
Dividing the numerator and the factor 5^n + n by 5^n gives
Since \displaystyle \frac{n}{5^n} \to 0, the numerator tends to 5 while the denominator grows without bound, so the ratio tends to 0 < 1. The series converges.
Splitting is shorter. The term is a sum of two terms,
using \displaystyle \frac{n}{n!} = \frac{1}{(n-1)!}. Each piece gives a convergent series by the ratio test in one line (ratios \displaystyle \frac{5}{n+1} and \displaystyle \frac1n, both tending to 0), and the sum of two convergent series converges.
It also leads to the sum, through a result quoted here without proof from a later lesson, The common Maclaurin series: \displaystyle \sum_{m=0}^{\infty} \frac{x^m}{m!} = e^x for every x. With it, the first piece is e^5 - 1 (the m = 0 term is missing) and the second is \displaystyle \sum_{m=0}^{\infty} \frac{1}{m!} = e. So
The ratio test gives the verdict only.
Day 1: Saia’s three candidates
Saia considers \sum_{n=1}^{\infty} n^2 e^{-n^3}. Compare the integral test, the ratio test and direct comparison on it.
SolutionThe term is f(n) with f(x) = x^2 e^{-x^3}, a function times the derivative of its exponent (up to a constant), so the integral test is available. The function is positive and continuous. For decrease,
which is negative for x \ge 1, since 3x^3 \ge 3 > 2. With u = x^3,
The integral converges, so the series converges. That took a derivative, a sign argument and a substitution.
The ratio test needs one line. Since (n+1)^3 - n^3 = 3n^2 + 3n + 1,
and 0 < 1. Same verdict, no hypotheses to check beyond a_n \ne 0.
Direct comparison with e^{-n} also works, since n^2 e^{-n^3} \le e^{-n} for n \ge 1, but proving that inequality takes longer than the ratio test. The terms also show how fast: a_1 = e^{-1} \approx 0.36788 and a_2 = 4e^{-8} \approx 0.00134, while a_3 = 9e^{-27} is about 1.7 \times 10^{-11}, so the sum is 0.36922 to five decimal places.
Day 2: factorials that cancel
Decide whether \sum_{n=1}^{\infty} \dfrac{n!}{(n+2)!} converges, and find its sum if it does.
SolutionFactorials suggest the ratio test, and it applies, since every term is positive:
The test is silent. The reason is visible before computing anything: the factorials cancel inside each term, since (n+2)! = (n+2)(n+1)\,n!, so
a rational function of n, on which the ratio test always gives 1. A factorial in the term is a reason for the ratio test only when it does not cancel.
The simplified term behaves like \displaystyle \frac{1}{n^2}, and limit comparison settles convergence: \displaystyle \frac{a_n}{1/n^2} = \frac{n^2}{(n+1)(n+2)} \to 1, and \displaystyle \sum \frac{1}{n^2} converges. A better route gives the sum as well. By partial fractions, \displaystyle \frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}, so the partial sums telescope:
The series converges, with sum \displaystyle \frac12.
Day 2: a term between the p series
Decide whether \displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n\sqrt{\ln n}} converges, after showing why the ratio test and each natural comparison say nothing.
SolutionThe term is \displaystyle \frac1n divided by a slowly growing factor. The ratio test gives
because both factors tend to 1 (for the second, \displaystyle \ln(n+1) - \ln n = \ln\left(1 + \frac1n\right) \to 0 while \ln n \to \infty). Silent, as it must be for a term made of powers and logarithms.
Comparison with \displaystyle \frac1n: for n \ge 3, \ln n > 1, so \displaystyle a_n < \frac1n. The term is smaller than a divergent series, which proves nothing. Limit comparison with \displaystyle \frac1n gives \displaystyle \frac{a_n}{1/n} = \frac{1}{\sqrt{\ln n}} \to 0, the edge case that decides nothing when \sum b_n diverges.
Comparison with a convergent \displaystyle \frac{1}{n^p}, p > 1: \displaystyle \frac{a_n}{1/n^p} = \frac{n^{p-1}}{\sqrt{\ln n}} \to \infty, since a power of n outgrows any power of \ln n. So a_n is eventually larger than a convergent series, which also proves nothing. The term sits in the gap between \displaystyle \frac1n and every \displaystyle \frac{1}{n^p} with p > 1, where no p series can serve as a benchmark.
The integral test can decide inside that gap. The function \displaystyle f(x) = \frac{1}{x\sqrt{\ln x}} is positive and continuous on [2, \infty), and decreasing there because its denominator is a product of positive increasing factors. With u = \ln x, \displaystyle du = \frac{dx}{x},
which tends to \infty as b \to \infty. The integral diverges, so the series diverges. The logarithm in the term together with its derivative \displaystyle \frac1x was the sign that the integral test would work.
Day 2: sorting absolute and conditional
Classify each series as absolutely convergent, conditionally convergent, or divergent: (a) \displaystyle \sum_{n=1}^{\infty} (-1)^n\,\dfrac{\arctan n}{\sqrt{n}}; (b) \displaystyle \sum_{n=1}^{\infty} (-1)^n\,\dfrac{\arctan n}{n\sqrt{n}}.
SolutionThe signs alternate, so begin with the absolute values. In both, \arctan n is a bounded factor: it increases with n and lies between \displaystyle \arctan 1 = \frac{\pi}{4} and \displaystyle \frac{\pi}{2}. Bounded factors are handled by direct comparison.
(b) \left\lvert a_n \right\rvert = \dfrac{\arctan n}{n^{3/2}} \le \dfrac{\pi/2}{n^{3/2}}, and \displaystyle \sum \frac{1}{n^{3/2}} converges (\displaystyle p = \frac32). So \sum \lvert a_n \rvert converges and the series converges absolutely. Nothing else is needed.
(a) \displaystyle \left\lvert a_n \right\rvert = \dfrac{\arctan n}{\sqrt n} \ge \dfrac{\pi/4}{\sqrt n}, and \displaystyle \sum \frac{1}{\sqrt n} diverges (\displaystyle p = \frac12). So the series does not converge absolutely. The terms do tend to 0, since the numerator is bounded and \displaystyle \sqrt n \to \infty, so the nth term test is silent and the alternating series test is next.
Its remaining hypothesis is that \displaystyle b_n = \frac{\arctan n}{\sqrt n} eventually decreases. With \displaystyle f(x) = \frac{\arctan x}{\sqrt x},
For x \ge 2, \displaystyle \frac{2x}{1+x^2} \le \frac45 (this fraction decreases for x \ge 1), while \arctan x \ge \arctan 2 \approx 1.107, so f'(x) < 0 and the b_n decrease from n = 2 on. Directly, \displaystyle b_1 = \frac{\pi}{4} \approx 0.7854 and \displaystyle b_2 = \frac{\arctan 2}{\sqrt 2} \approx 0.7829, so b_1 > b_2 as well. (The function f itself rises briefly between x = 1 and about x = 1.39; the test needs the sequence to decrease, not the function.) The alternating series test applies, and series (a) converges conditionally.
Boundaries
- Every test has hypotheses. Comparison, limit comparison and the integral test need positive terms (or are applied to \lvert a_n \rvert); the integral test also needs f continuous and eventually decreasing; the alternating series test needs b_n eventually decreasing to 0; the ratio test needs a_n \ne 0. A test whose hypotheses fail gives no information, even if its computation can be carried out.
- A constant is needed for a verdict by form. The geometric and p series tests need a fixed ratio and a fixed exponent. A drifting exponent, as in \displaystyle \frac{1}{n^{1+1/\ln n}}, needs another argument even when every exponent is above 1 (Example 2).
- Silence is not a verdict. L = 1 in the ratio or root test, an inequality in the wrong direction, a limit comparison limit of 0 or \infty on the wrong side, and a_n \to 0 in the nth term test all leave the question open. None of them suggests convergence or divergence.
- The alternating series test proves only conditional-or-better convergence. It cannot show divergence, and it cannot show absolute convergence; for those, test \sum \lvert a_n \rvert.
- The choice of test never changes the answer. Every test that applies gives the same verdict; choosing well changes only the length of the argument.
Where this goes next
A power series \sum c_n (x - a)^n is settled by the ratio or root test for all x except at the ends of an interval, where the limit is exactly 1. At those two points the series becomes an ordinary numerical series of whatever shape the coefficients give, and Power series and the interval of convergence uses exactly the reading of this lesson to decide them, often with one endpoint conditionally convergent and the other divergent. The sums in Examples 3 and 5 also point ahead: once a series is known to converge, Taylor and Maclaurin series and telescoping can sometimes give its exact value.
Exercises
Day 1: read the shape, then settle
Decide whether each series converges or diverges. Name the test you chose and say what in the term suggested it.
- \displaystyle \sum_{n=1}^{\infty} \dfrac{3n^2 - 1}{n^3\sqrt{n} + 2}
- \sum_{n=1}^{\infty} \dfrac{(n+3)!}{n!\,4^n}
- \sum_{n=1}^{\infty} \left(\dfrac{2n+1}{3n-1}\right)^n
- \sum_{n=1}^{\infty} \dfrac{n!}{10^n}
- \sum_{n=1}^{\infty} \dfrac{1}{2 + \sin n}
- \sum_{n=1}^{\infty} \dfrac{3^n + (-2)^n}{5^n} (find the sum if it converges)
- \sum_{n=2}^{\infty} \dfrac{1}{n^{\ln n}}
- \displaystyle \sum_{n=1}^{\infty} \dfrac{\sqrt{n} + \sin n}{n^2}
- \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n} + \ln n}
- \sum_{n=1}^{\infty} \ln\left(1 + \dfrac{1}{n^2}\right)
Day 1: two routes
- Show that \sum_{n=1}^{\infty} \dfrac{n}{(n^2+1)^2} converges by the integral test and again by limit comparison. Which route is shorter, and why?
- Decide whether \sum_{n=1}^{\infty} \left(\dfrac{n}{n+2}\right)^{n^2} converges by the root test, then explain why the ratio test is the longer route here.
Day 1: reasoning
- Let P and Q be polynomials of degrees p and q with \displaystyle \frac{P(n)}{Q(n)} > 0 for every n \ge 1. Explain which test settles \displaystyle \sum \frac{P(n)}{Q(n)}, and show the series converges exactly when q \ge p + 2.
- Saia says the integral test should always be tried first, because it works on every series with positive terms. Give two shapes of term for which it cannot be used or is impractical, and say what to use instead.
Day 2: when the first test is silent
- Show that the ratio test is silent on \displaystyle \sum_{n=1}^{\infty} \dfrac{\sqrt{n} + 1}{n^2 - n + 1}, then decide convergence.
- Decide whether \sum_{n=2}^{\infty} \dfrac{1}{(\ln n)^3} converges. Direct comparison with \displaystyle \frac1n works only from some index on; find that index.
- Decide whether \sum_{n=1}^{\infty} \dfrac{1}{n^{2 - 1/n}} converges.
- Malakai compares \displaystyle \sum_{n=2}^{\infty} \dfrac{\sqrt{\ln n}}{n^2} with \displaystyle \sum \frac{1}{n^2}. Explain why that comparison decides nothing, then settle the series.
Day 2: absolute or conditional
Classify each series as absolutely convergent, conditionally convergent, or divergent.
- \sum_{n=1}^{\infty} (-1)^n\,\dfrac{n^2}{n^3 + 10} (give the index from which the alternating series test’s terms decrease)
- \sum_{n=1}^{\infty} (-1)^n\,\dfrac{n!}{(2n)!}
- \sum_{n=1}^{\infty} (-1)^n\left(\dfrac{n+1}{n+3}\right)^n
- \displaystyle \sum_{n=1}^{\infty} (-1)^n\,\dfrac{\sqrt{n}}{2n - 1}
Day 2: reasoning
- Haukea says: “If the ratio test gives L = 1, the series converges conditionally.” Give two counterexamples of different kinds.
- Explain why neither the ratio test nor the root test can ever show that a series converges conditionally.
- Suppose a_n > 0 and \sum a_n converges. Which test shows that \sum \sin(a_n) converges? Give the argument.
Answers
- Converges, by limit comparison with \displaystyle \frac{1}{n^{3/2}}: keeping the dominant pieces gives \displaystyle \frac{3n^2}{n^{7/2}}, and \displaystyle \frac{(3n^2 - 1)\,n^{3/2}}{n^{7/2} + 2} \to 3; \displaystyle \sum \frac{1}{n^{3/2}} converges.
- Converges, by the ratio test: the factorials leave (n+1)(n+2)(n+3), a polynomial against 4^n, and \displaystyle \frac{a_{n+1}}{a_n} = \frac{n+4}{4(n+1)} \to \frac14 < 1.
- Converges, by the root test, since the whole term is raised to the nth power: \displaystyle \sqrt[n]{a_n} = \frac{2n+1}{3n-1} \to \frac23 < 1.
- Diverges. The factorial outgrows 10^n, so a_n \to \infty and the nth term test applies; the ratio \displaystyle \frac{n+1}{10} \to \infty says the same.
- Diverges by the nth term test: 2 + \sin n \le 3, so \displaystyle a_n \ge \frac13 for every n, and the terms do not tend to 0 (they have no limit at all).
- Converges, with sum \displaystyle \frac{17}{14}. It is the sum of two geometric series: \displaystyle \sum_{n=1}^{\infty} \left(\frac35\right)^n = \frac{3/5}{2/5} = \frac32 and \displaystyle \sum_{n=1}^{\infty} \left(-\frac25\right)^n = \frac{-2/5}{7/5} = -\frac27; \displaystyle \frac32 - \frac27 = \frac{17}{14}.
- Converges by direct comparison: the exponent \ln n grows, so for n \ge 8 (where \ln n \ge \ln 8 \approx 2.08 > 2) n^{\ln n} \ge n^2 and \displaystyle a_n \le \frac{1}{n^2}. The first six terms do not affect convergence.
- Converges by direct comparison, since \sin n is a bounded factor: \displaystyle 0 < \sqrt n + \sin n \le 2\sqrt n for n \ge 1, so \displaystyle 0 < a_n \le \frac{2}{n^{3/2}}, a convergent p series times 2.
- Diverges by direct comparison: \displaystyle \ln n \le \sqrt n for every n \ge 1 (the function \displaystyle \frac{\ln x}{\sqrt x} has maximum \displaystyle \frac{2}{e} < 1 at x = e^2), so \displaystyle a_n \ge \frac{1}{2\sqrt n}, and \displaystyle \sum \frac{1}{\sqrt n} diverges.
- Converges, by limit comparison with \displaystyle \frac{1}{n^2}: with \displaystyle t = \frac{1}{n^2} \to 0, \displaystyle \frac{\ln(1 + t)}{t} \to 1, and \displaystyle \sum \frac{1}{n^2} converges.
- Converges. Integral test: \displaystyle f(x) = \frac{x}{(x^2+1)^2} is positive and continuous, \displaystyle f'(x) = \frac{1 - 3x^2}{(x^2+1)^3} < 0 for x \ge 1, and \displaystyle \int_1^{\infty} f(x)\,dx = \left[-\frac{1}{2(x^2+1)}\right]_1^{\infty} = \frac14. Limit comparison with \displaystyle \frac{1}{n^3}: \displaystyle \frac{n^4}{(n^2+1)^2} \to 1. Limit comparison is shorter: one limit, with no derivative to check and no antiderivative to find. The integral route does give more, the bounds \displaystyle \frac14 < S < \frac12 (the integral, and the integral plus \displaystyle a_1 = \frac14).
- Converges. The whole term is raised to the power n^2, so the root test undoes most of it: \displaystyle \sqrt[n]{a_n} = \left(\frac{n}{n+2}\right)^n = \left(1 + \frac2n\right)^{-n} \to e^{-2} < 1. In the ratio \displaystyle \frac{a_{n+1}}{a_n} the bases differ and the exponents (n+1)^2 and n^2 differ, so nothing cancels, and finding its limit means taking logarithms and expanding; the root test needed one standard limit.
- Keeping leading terms, \displaystyle \frac{P(n)}{Q(n)} behaves like \displaystyle \frac{c}{n^{q-p}}, c > 0 the ratio of the leading coefficients (positive, since the terms are positive). Limit comparison with \displaystyle \frac{1}{n^{q-p}} gives the limit c, so the series behaves like the p series with exponent q - p, which converges exactly when q - p > 1, that is q \ge p + 2 for integer degrees. The ratio and root tests give 1 here and are no help.
- Any two of: a factorial, as in \displaystyle \frac{2^n}{n!}, has no continuous f that is easy to integrate (use the ratio test); a bounded oscillating factor, as in \displaystyle \frac{2 + \cos n}{n^2}, gives an f that is not decreasing (use direct comparison); \displaystyle \frac{1}{n^{\ln n}} has no usable antiderivative (use direct comparison); an alternating term is not positive (test \lvert a_n \rvert, or use the alternating series test).
- Converges. The ratio is the product of \displaystyle \frac{\sqrt{n+1} + 1}{\sqrt n + 1} and \displaystyle \frac{n^2 - n + 1}{n^2 + n + 1}, and both factors tend to 1. Limit comparison with \displaystyle \frac{1}{n^{3/2}}: the quotient is \displaystyle \frac{(\sqrt n + 1)\,n^{3/2}}{n^2 - n + 1}, that is \displaystyle \frac{n^2 + n^{3/2}}{n^2 - n + 1}, which tends to 1.
- Diverges; \displaystyle a_n > \frac1n for every n \ge 94. Limit comparison with \displaystyle \frac1n gives \displaystyle \frac{n}{(\ln n)^3} \to \infty, and since \displaystyle \sum \frac1n diverges this edge case gives divergence. For direct comparison, (\ln n)^3 < n fails from n = 7 to n = 93 (for instance (\ln 93)^3 \approx 93.1) and holds for n \ge 94 ((\ln 94)^3 \approx 93.8).
- Converges. \displaystyle a_n = \frac{\sqrt[n]{n}}{n^2}, and limit comparison with \displaystyle \frac{1}{n^2} gives \displaystyle \sqrt[n]{n} \to 1. (Directly, \displaystyle \sqrt[n]{n} \le \sqrt[3]{3} < \frac32 for every n, the largest value being at n = 3, so \displaystyle a_n < \frac{3}{2n^2}.)
- Converges. For n \ge 3, \displaystyle \sqrt{\ln n} > 1, so \displaystyle a_n > \frac{1}{n^2}: larger than a convergent series, which proves nothing. Instead, \ln n < n, so \displaystyle \sqrt{\ln n} < \sqrt n and \displaystyle a_n < \frac{\sqrt n}{n^2} = \frac{1}{n^{3/2}}, a convergent p series.
- Conditionally convergent; the terms decrease from n = 3. \lvert a_n \rvert behaves like \displaystyle \frac1n: \displaystyle \frac{n^3}{n^3 + 10} \to 1, so \sum \lvert a_n \rvert diverges. With \displaystyle f(x) = \frac{x^2}{x^3 + 10}, \displaystyle f'(x) = \frac{x(20 - x^3)}{(x^3+10)^2} < 0 for \displaystyle x > \sqrt[3]{20} \approx 2.71; the values \displaystyle b_1 = \frac{1}{11}, \displaystyle b_2 = \frac29, \displaystyle b_3 = \frac{9}{37}, \displaystyle b_4 = \frac{8}{37} rise to b_3 and fall after. b_n \to 0, so the alternating series test applies.
- Absolutely convergent, by the ratio test on \lvert a_n \rvert: the ratio is \displaystyle \frac{(n+1)!}{(2n+2)!}\cdot\frac{(2n)!}{n!}, which cancels to \displaystyle \frac{n+1}{(2n+2)(2n+1)} = \frac{1}{2(2n+1)} \to 0.
- Divergent. \displaystyle \lvert a_n \rvert = \left(1 + \frac{2}{n+1}\right)^{-n} \to e^{-2} \ne 0, so the nth term test applies (the signs alternate, but the alternating series test needs b_n \to 0). The root test is silent: \displaystyle \sqrt[n]{\lvert a_n \rvert} = \frac{n+1}{n+3} \to 1.
- Conditionally convergent. \displaystyle \frac{\sqrt n}{2n - 1} \ge \frac{\sqrt n}{2n} = \frac{1}{2\sqrt n}, so \sum \lvert a_n \rvert diverges. With \displaystyle f(x) = \frac{\sqrt x}{2x - 1}, \displaystyle f'(x) = -\frac{2x + 1}{2\sqrt x\,(2x-1)^2} < 0 for x \ge 1, and b_n \to 0, so the alternating series test applies.
- \displaystyle \sum \frac{1}{n^2} has L = 1 and converges absolutely (its terms are positive); \displaystyle \sum \frac1n has L = 1 and diverges. L = 1 allows every outcome, including conditional convergence, as for \displaystyle \sum \frac{(-1)^n}{n}.
- If L < 1, the test proves that \sum \lvert a_n \rvert converges, so the convergence is absolute. If L > 1, the terms do not tend to 0 and the series diverges. So the only conclusions the tests can reach are absolute convergence and divergence; a conditionally convergent series always gives L = 1 or no limit.
- Direct comparison. Since \sum a_n converges, a_n \to 0, so from some N on 0 < a_n < \pi, where 0 < \sin(a_n) \le a_n (from \sin x \le x for x \ge 0). So \sum_{n \ge N} \sin(a_n) has positive terms below those of a convergent series and converges; the finitely many terms before N do not affect convergence.