The integral test and p series
The terms of a positive series often have no formula for their partial sums. When the terms are the values of a decreasing function, can the area under its graph decide whether the series has a sum, and say how close a partial sum comes to it?
Where you are 6 in Convergence tests
Why this exists
Pekka sets up an idealized row of identical loudspeakers along a straight line, one every 3 meters, the first 3 meters from where Aatos stands, the row running on without end. A single speaker delivers sound intensity \displaystyle \frac{k}{d^2} at distance d (the inverse square law for a point source), so the total intensity reaching Aatos is
Is it finite? The terms tend to 0, so the The nth term test for divergence is silent, and \displaystyle \frac{1}{n^2} does not telescope the way the series in Infinite series and partial sums do. Had the speakers been long line arrays, whose sound fades like \displaystyle \frac{1}{d}, the total would be a multiple of the harmonic series and infinite.
The graph of \displaystyle y = \frac{1}{x^2} passes through every term, and the area under it out to infinity is an improper integral already known to converge. The integral test turns such an area into a verdict on the series, and it settles every series \displaystyle \sum \frac{1}{n^p} at once.
Definition
In plain words: when the terms of a series are the values at whole numbers of a positive, decreasing function, the series and the area under the graph are either both finite or both infinite.
The integral test. Let m be an integer and let f be continuous, positive and decreasing on [m, \infty), with a_n = f(n) for every integer n \ge m. Then \sum_{n=m}^{\infty} a_n and \int_m^{\infty} f(x)\,dx both converge or both diverge.
The hypotheses need only hold eventually. Adding or removing finitely many terms never changes whether a series converges, so \sum_{n=1}^{\infty} a_n is settled by the test applied from any m beyond which f qualifies.
p-series. For a real number p, the series \sum_{n=1}^{\infty} \dfrac{1}{n^p} is a p-series. At p = 1 it is the harmonic series.
Remainder. If \sum_{n=1}^{\infty} a_n converges to S, the remainder after N terms is
the error made when the partial sum S_N stands in for S.
Intuition
Day 1: rectangles under and over a curve
On [n, n+1] a decreasing f stays between a_{n+1} and a_n. A rectangle of width 1 and height a_n has area a_n, so a sum of terms is an area made of rectangles.
In Figure 1 the rectangle on [n, n+1] has height a_n, the value at its left end, and covers the region under the curve there. So S_N \ge \int_1^{N+1} f(x)\,dx: if the integral grows without bound, it drags the partial sums up with it.
In Figure 2 each rectangle moves one step left, onto [n-1, n], and now fits under the curve. So a_2 + \cdots + a_N \le \int_1^N f(x)\,dx: if the integral converges, the partial sums are bounded, and since the terms are positive the partial sums increase, so they settle.
Day 2: the tail is an area too
The same two pictures, started at N instead of 1, trap the remainder R_N = a_{N+1} + a_{N+2} + \cdots between the area to the right of N + 1 and the area to the right of N. A partial sum together with two improper integrals pins the sum in a short interval.
Results
The integral test, as stated in the Definition.
Bounds on partial sums. Under the hypotheses of the integral test, for every N \ge m,
Bounds on the sum. If the integral converges,
The p-series test. \sum_{n=1}^{\infty} \dfrac{1}{n^p} converges if p > 1 and diverges if p \le 1.
Remainder bounds. If f satisfies the hypotheses on [N, \infty) and the series converges,
For a p-series with p > 1 this reads
So S lies between S_N + \int_{N+1}^{\infty} f and S_N + \int_N^{\infty} f, and the midpoint of that interval estimates S with an error of at most half the interval’s width.
Why it is true
The rectangles. For n \ge m and n \le x \le n + 1, f decreasing gives a_{n+1} \le f(x) \le a_n. Integrating over an interval of width 1 (the comparison property of definite integrals),
Adding the right-hand inequality for n = m, \ldots, N and using additivity over intervals gives \int_m^{N+1} f \le \sum_{n=m}^{N} a_n. Adding the left-hand one for n = m, \ldots, N - 1 gives \sum_{n=m+1}^{N} a_n \le \int_m^N f; add a_m to both sides.
The test. The partial sums T_N = \sum_{n=m}^{N} a_n increase with N, because every term is positive. If the integral converges, then T_N \le a_m + \int_m^N f \le a_m + \int_m^{\infty} f, the last step because f > 0. An increasing sequence that is bounded above has a limit: the completeness of the real numbers, the same fact behind comparison for integrals, taken here without proof. If the integral diverges, then \int_m^b f \to \infty (a positive integrand cannot make it oscillate), and T_N \ge \int_m^{N+1} f follows it to infinity. Bounds on the sum come from letting N \to \infty.
Remainders. Adding the left-hand inequality for n = N, \ldots, K - 1 gives a_{N+1} + \cdots + a_K \le \int_N^K f, and the right-hand one for n = N + 1, \ldots, K gives \int_{N+1}^{K+1} f \le a_{N+1} + \cdots + a_K. Let K \to \infty.
The p-series. If p \le 0, then \displaystyle \frac{1}{n^p} = n^{-p} \ge 1, the terms do not tend to 0, and the series diverges. If p > 0, then f(x) = x^{-p} is continuous and positive on [1, \infty) and decreasing, since f'(x) = -p\,x^{-p-1} < 0. By the integral test the series behaves as \int_1^{\infty} x^{-p}\,dx does, and that converges exactly when p > 1 (Improper integrals). For p > 1, \displaystyle \int_N^{\infty} x^{-p}\,dx = \frac{N^{1-p}}{p-1}, which gives the remainder bounds.
Worked examples
Day 1: a substitution, and terms that rise before they fall
Decide whether \sum_{n=1}^{\infty} \dfrac{n}{(n^2+5)^{3/2}} converges, and if it does, find an interval containing its sum.
SolutionLet f(x) = x(x^2+5)^{-3/2}, continuous and positive for x \ge 1. By the product and chain rules,
which is negative only once \displaystyle x^2 > \frac{5}{2}, that is \displaystyle x > \sqrt{5/2} \approx 1.58. Indeed \displaystyle a_1 = \frac{1}{6\sqrt{6}} \approx 0.0680 is less than \displaystyle a_2 = \frac{2}{27} \approx 0.0741. So f is decreasing on [2, \infty) and the test applies with m = 2. With u = x^2 + 5, du = 2x\,dx,
The integral converges, so \sum_{n=2}^{\infty} a_n converges, and adding the single term a_1 keeps it convergent. The bounds on the sum with m = 2 give \displaystyle \frac{1}{3} \le \sum_{n=2}^{\infty} a_n \le \frac{2}{27} + \frac{1}{3}, and adding a_1,
A machine sum gives about 0.4395. The bounds cannot start at m = 1: on [1, 2] the function rises, and the inequality a_{n+1} \le \int_n^{n+1} f \le a_n behind them fails.
Day 1: two series that crawl
Decide whether each converges: (a) \sum_{n=2}^{\infty} \dfrac{1}{n(\ln n)^{3/2}}, (b) \sum_{n=3}^{\infty} \dfrac{1}{n\ln n\,\ln(\ln n)}.
SolutionEach denominator is a product of positive increasing factors on the range used (\ln x > 0 for x > 1, \ln(\ln x) > 0 for x > e), so each f is continuous, positive and decreasing. Series (b) starts at 3 because \ln(\ln 2) < 0.
(a) With u = \ln x, \displaystyle du = \frac{dx}{x},
It converges. With \displaystyle a_2 = \frac{1}{2(\ln 2)^{3/2}} \approx 0.866, the sum lies between 2.402 and 3.269.
(b) With u = \ln(\ln x), \displaystyle du = \frac{dx}{x\ln x},
The logarithm grows without bound, so three of them in a row do too: the integral diverges, and so does the series. How slowly? By the upper bound on partial sums, S_N \le a_3 + \ln(\ln(\ln N)) - \ln(\ln(\ln 3)), with a_3 \approx 3.226 and -\ln(\ln(\ln 3)) \approx 2.364. At N = 10^{100}, \ln N \approx 230.26, \ln(\ln N) \approx 5.439 and \ln(\ln(\ln N)) \approx 1.694, so the partial sum is less than 7.29. After 10^{100} terms the total is still under 7.29, and yet it passes every number eventually.
Series (a) differs from the harmonic series only by the factor (\ln n)^{3/2}, which grows more slowly than any positive power of n; that is still enough to make it converge.
Day 1: when “decreasing” fails
Let f(x) = \dfrac{1}{x^2} + \sin^2(\pi x). Compare \sum_{n=1}^{\infty} f(n) with \int_1^{\infty} f(x)\,dx.
SolutionAt every integer \sin(\pi n) = 0, so \displaystyle f(n) = \frac{1}{n^2} and the series is \displaystyle \sum \frac{1}{n^2}. The integral test applied to \displaystyle \frac{1}{x^2} (which does qualify) bounds its partial sums by 1 + \int_1^{\infty} x^{-2}\,dx = 2, so it converges.
The integral diverges. Since \displaystyle \sin^2(\pi x) = \frac{1 - \cos(2\pi x)}{2},
which grows without bound, while the \displaystyle \frac{1}{x^2} part adds only a finite amount.
Here f is continuous and positive but not decreasing: between every two integers it climbs to about 1 and comes back. The integral counts the area of those bumps; the series samples f only at the integers, where the bumps vanish. Without “decreasing” the conclusion of the test is false. Exercise 11 builds a function that fails the other way.
Day 2: reading the exponent
Decide whether each converges: (a) \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{\sqrt[3]{n^4}}, with bounds on its sum; (b) \sum_{n=1}^{\infty} \dfrac{1}{n^{0.98}}; (c) \displaystyle \sum_{n=1}^{\infty} \dfrac{\sqrt{n}}{n\sqrt[3]{n^2}}.
Solution(a) \displaystyle \sqrt[3]{n^4} = n^{4/3}, so \displaystyle p = \frac{4}{3} > 1 and the series converges. With m = 1, \displaystyle \int_1^{\infty} x^{-4/3}\,dx = \frac{1}{p - 1} = 3, so the bounds on the sum give 3 \le S \le 4. (A machine sum gives about 3.6009.)
(b) p = 0.98 \le 1, so it diverges. The exponents 0.98 and 1.02 are close, but the dividing line sits exactly at 1, and the two series fall on opposite sides of it.
(c) Combine the powers: \displaystyle \frac{n^{1/2}}{n \cdot n^{2/3}} = n^{1/2 - 5/3} = \frac{1}{n^{7/6}}. Here \displaystyle p = \frac{7}{6} > 1, so it converges.
Day 2: Pekka’s loudspeakers
Return to the row of speakers in “Why this exists”, with total intensity \displaystyle \frac{k}{9}\sum_{n=1}^{\infty} \frac{1}{n^2}. (a) Is the total finite? (b) Taking \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} (a result of Euler’s, stated here without proof), find the least number N of speakers whose sound is at least 99\% of the endless row’s total.
Solution(a) p = 2 > 1, so the series converges and the total is \displaystyle \frac{k}{9} \cdot \frac{\pi^2}{6} = \frac{\pi^2 k}{54}.
(b) The speakers beyond the Nth contribute \displaystyle \frac{k}{9}R_N, where R_N is the remainder of \displaystyle \sum \frac{1}{n^2}. The factor \displaystyle \frac{k}{9} cancels from both sides, so the answer is the same for any speaker and any spacing. The condition is
For p = 2 the remainder bounds are \displaystyle \frac{1}{N+1} \le R_N \le \frac{1}{N}. The upper bound meets the target once \displaystyle N \ge \frac{1}{0.016449} \approx 60.8, so 61 speakers are enough. The lower bound rules out 59, and so every smaller N, since R_N shrinks as N grows: \displaystyle R_{59} \ge \frac{1}{60} \approx 0.016667, above the target. For N = 60 the bounds straddle the target (\displaystyle \frac{1}{61} \approx 0.016393 and \displaystyle \frac{1}{60}), so compute: \displaystyle R_{60} = \frac{\pi^2}{6} - S_{60} \approx 0.016529, still above. The least number is 61.
Day 2: an estimate with a guaranteed error
Estimate \sum_{n=1}^{\infty} \dfrac{1}{n^{5/2}} from S_{10} and the remainder bounds. How many terms would the partial sum alone need to be within 10^{-4} of the sum?
SolutionWith \displaystyle p = \frac{5}{2}, \displaystyle p - 1 = \frac{3}{2}, and the remainder bounds read
Adding ten terms, S_{10} \approx 1.321921. At N = 10 the bounds are \displaystyle R_{10} \ge \frac{2}{3 \cdot 11^{3/2}} \approx 0.018273 and \displaystyle R_{10} \le \frac{2}{3 \cdot 10^{3/2}} \approx 0.021082, so
The midpoint 1.341598 is within half the width, about 0.0014, of S. The partial sum alone is only known to be within 0.0211.
For the partial sum alone to be within 10^{-4}, it suffices that \displaystyle \frac{2}{3N^{3/2}} \le 10^{-4}, that is \displaystyle N^{3/2} \ge \frac{20{,}000}{3}, or N \ge 354.2. So 355 terms guarantee it. (The sum is 1.341487 to six decimals, so the midpoint after ten terms is in fact off by about 0.0001.)
Day 2: measuring a divergent p-series
The series \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{\sqrt[3]{n}} diverges, since \displaystyle p = \frac{1}{3}. Use the bounds on partial sums to bracket the first N for which S_N > 100.
SolutionWith f(x) = x^{-1/3}, \displaystyle \int_1^N f(x)\,dx = \frac{3}{2}\left(N^{2/3} - 1\right), and the bounds with m = 1 read
For S_N > 100 the upper bound must exceed 100: N^{2/3} > 67, so N > 67^{3/2} \approx 548.4, and at least 549 terms are needed. The lower bound exceeds 100 once \displaystyle (N+1)^{2/3} > \frac{203}{3}, that is N + 1 > 556.6, so N = 556 certainly works. Adding the terms directly, S_{551} \approx 99.903 and S_{552} \approx 100.025: the first such N is 552. The bounds also show how the series diverges: S_N grows like \displaystyle \frac{3}{2}N^{2/3}.
Boundaries
- All three hypotheses matter, from some point on. Continuity lets the integral exist; positivity makes the partial sums increase; “decreasing” ties the samples f(n) to the area. Example 3 drops “decreasing” and the conclusion fails. Each may fail on an initial stretch (Example 1) as long as it holds from some m on.
- It is a test for series of positive terms. For terms that change sign, see Absolute and conditional convergence and Alternating series and their error bound.
- A verdict, not a sum. \displaystyle \sum \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645, while \int_1^{\infty} x^{-2}\,dx = 1. The integral locates the sum only within the bounds.
- It needs an f that can be integrated. Terms such as \displaystyle \frac{n!}{5^n} come from no function with a usable antiderivative; the The ratio and root tests is the tool there.
- The p-series test is about \displaystyle \frac{1}{n^p} exactly. A constant factor and finitely many terms change nothing; a series that only resembles a p-series is handled by comparison.
- The remainder bounds need f decreasing from N on, and they bound the error of S_N, not of any other estimate.
Where this goes next
The p-series are the yardsticks of the subject: the The direct and limit comparison tests decides a series such as \displaystyle \sum \frac{n+1}{n^3 + 2} by setting it beside \displaystyle \sum \frac{1}{n^2}. Recognizing a p-series or something close to one is a first step in Choosing a convergence test. The remainder bounds are one of two error estimates in the subject; the other, for alternating series, is in Alternating series and their error bound. And p-series return at the endpoints of power series, where a series like \displaystyle \sum \frac{x^n}{n} becomes the harmonic series at x = 1 (Power series and the interval of convergence).
Exercises
Day 1: the integral test
Decide whether the series converges. Check the hypotheses, and give the value of the integral you use.
- \sum_{n=1}^{\infty} \dfrac{1}{(3n+2)^2}
- \sum_{n=1}^{\infty} \dfrac{n^2}{n^3 + 4}
- \displaystyle \sum_{n=2}^{\infty} \dfrac{\sqrt[3]{\ln n}}{n}
- \sum_{n=1}^{\infty} \dfrac{1}{n^2 + 4n + 8}
- \sum_{n=1}^{\infty} n e^{-n/3}
- \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}\,(1 + \sqrt{n})^2}
- \sum_{n=3}^{\infty} \dfrac{1}{n\ln n\,(\ln(\ln n))^2}
Day 1: bounds
- Show that the sum S of the series in Exercise 4 satisfies 0.29 < S < 0.38.
- Use the bounds on partial sums to find an interval containing \sum_{n=1}^{100} \dfrac{n^2}{n^3 + 4}. (Mind where the function starts to decrease.)
Day 1: reasoning
- Explain why the integral test cannot be applied to f(x) = \dfrac{2 + \cos(\pi x)}{x^2}, then show that \sum_{n=1}^{\infty} \dfrac{2 + \cos(\pi n)}{n^2} converges anyway.
- Describe a continuous, positive function f on [1, \infty) for which \int_1^{\infty} f(x)\,dx converges but \sum_{n=1}^{\infty} f(n) diverges.
- Let f be continuous, positive and decreasing on [1, \infty), with a_n = f(n), and let D_N = \sum_{n=1}^{N} a_n - \int_1^N f(x)\,dx. Show that D_N decreases, that D_N \ge 0, and conclude that D_N has a limit.
Day 2: p-series
Decide whether the series converges.
- \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^{\sqrt{2}}}
- \displaystyle \sum_{n=1}^{\infty} \dfrac{\sqrt[4]{n}}{n}
- \sum_{n=1}^{\infty} 2^{-\ln n}
- \sum_{n=1}^{\infty} 3^{-\ln n}
- \displaystyle \sum_{n=1}^{\infty} \dfrac{5}{\sqrt[5]{n^6}}
- For which real numbers p does \sum_{n=1}^{\infty} \dfrac{1}{n^{3 - 2p}} converge?
Day 2: remainders and estimates
- Find an interval containing \sum_{n=1}^{\infty} \dfrac{1}{n^4} from S_5 and the remainder bounds.
- How many terms of \sum_{n=1}^{\infty} \dfrac{1}{n^{3/2}} does the upper remainder bound require for the partial sum to be within 0.01 of the sum?
- Estimate \sum_{n=1}^{\infty} \dfrac{1}{n^3} by the midpoint of the interval given by S_4 and the remainder bounds, and state the guaranteed error.
- Sora models a row of speakers in which speaker n contributes \dfrac{k}{n^3} at her seat. Show that the speakers beyond the tenth contribute less than 0.5\% of the endless row’s total.
- Bracket the first N for which \displaystyle \sum_{n=1}^{N} \dfrac{1}{\sqrt{n}} > 50 using the bounds on partial sums.
Day 2: reasoning
- For p > 1 let S(p) be the sum of the p-series. Show that \dfrac{1}{p-1} \le S(p) \le \dfrac{p}{p-1}, and find the limit of (p - 1)S(p) as p \to 1^+.
- For 0 < p < 1, show that the partial sums of \sum \dfrac{1}{n^p} satisfy \lim_{N \to \infty} \dfrac{S_N}{N^{1-p}} = \dfrac{1}{1-p}.
Answers
- Converges. f(x) = (3x+2)^{-2} is positive and decreasing, and \displaystyle \int_1^b (3x+2)^{-2}\,dx = \frac{1}{15} - \frac{1}{3(3b+2)} \to \frac{1}{15}.
- Diverges. \displaystyle f'(x) = \frac{x(8 - x^3)}{(x^3+4)^2} < 0 for x > 2, so f decreases on [2, \infty), and \displaystyle \int_2^b f(x)\,dx = \frac{1}{3}\ln\frac{b^3 + 4}{12} \to \infty.
- Diverges. \displaystyle f(x) = \frac{(\ln x)^{1/3}}{x} has \displaystyle f'(x) = \frac{(\ln x)^{-2/3}\left(\frac{1}{3} - \ln x\right)}{x^2} < 0 for x > e^{1/3}, so it decreases on [2, \infty). With u = \ln x, \displaystyle \int_2^b f(x)\,dx = \frac{3}{4}\left((\ln b)^{4/3} - (\ln 2)^{4/3}\right) \to \infty.
- Converges. n^2 + 4n + 8 = (n+2)^2 + 4 is positive and increasing. The antiderivative is \displaystyle \frac{1}{2}\arctan\frac{x+2}{2}, so the integral from 1 to \infty is \displaystyle \frac{1}{2}\left(\frac{\pi}{2} - \arctan\frac{3}{2}\right), which is \displaystyle \frac{1}{2}\arctan\frac{2}{3} \approx 0.2940.
- Converges. \displaystyle f'(x) = e^{-x/3}\left(1 - \frac{x}{3}\right) < 0 for x > 3. By parts, \int x e^{-x/3}\,dx = -3xe^{-x/3} - 9e^{-x/3} + C, so \displaystyle \int_3^{\infty} x e^{-x/3}\,dx = \frac{18}{e} \approx 6.622.
- Converges. Both factors of the denominator increase. With \displaystyle u = 1 + \sqrt{x}, \displaystyle du = \frac{dx}{2\sqrt{x}}, \displaystyle \int_1^b f(x)\,dx = 1 - \frac{2}{1 + \sqrt{b}} \to 1.
- Converges. The denominator is a product of positive increasing factors for x \ge 3. With u = \ln(\ln x), \displaystyle \int_3^{\infty} f(x)\,dx = \frac{1}{\ln(\ln 3)} \approx 10.63.
- With m = 1: the integral is \approx 0.2940 (Exercise 4) and \displaystyle a_1 = \frac{1}{13} \approx 0.0769, so 0.2940 \le S \le 0.3710.
- Between 3.986 and 4.311. The function decreases only from 2 (Exercise 2), so bound a_2 + \cdots + a_{100} with m = 2. It is at least \displaystyle \frac{1}{3}\ln\frac{101^3+4}{12} and at most \displaystyle \frac{1}{3} + \frac{1}{3}\ln\frac{100^3+4}{12}; then add \displaystyle a_1 = \frac{1}{5}. (Directly, S_{100} \approx 4.149.)
- f is not decreasing: \cos(\pi x) climbs from -1 to 1 on every [2j - 1, 2j], and for instance \displaystyle f(3) = \frac{1}{9} < f(4) = \frac{3}{16}. But the terms are positive with \displaystyle a_n \le \frac{3}{n^2}, so S_N is at most 3 times a partial sum of \displaystyle \sum \frac{1}{n^2}. Those are at most 1 + \int_1^{\infty} x^{-2}\,dx = 2, so S_N \le 6. Increasing and bounded, the partial sums converge.
- Take \displaystyle f(x) = \frac{1}{x^2} plus, at each integer n \ge 2, a triangular spike of height 1 on \displaystyle \left[n - \frac{1}{n^2}, n + \frac{1}{n^2}\right]. The spikes do not overlap, each adds area \displaystyle \frac{1}{n^2}, so the integral is at most \displaystyle 1 + \sum_{n \ge 2} \frac{1}{n^2}, finite. But f(n) \ge 1, so the series diverges by the The nth term test for divergence.
- D_{N+1} - D_N = a_{N+1} - \int_N^{N+1} f \le 0 by the rectangle inequality. And D_N \ge \int_1^{N+1} f - \int_1^N f \ge 0 by the lower bound on partial sums and f > 0. A decreasing sequence bounded below has a limit. For \displaystyle f(x) = \frac{1}{x} the limit is the Euler constant of Infinite series and partial sums.
- Converges: \displaystyle p = \sqrt{2} \approx 1.414 > 1.
- Diverges: \displaystyle \frac{n^{1/4}}{n} = \frac{1}{n^{3/4}}, \displaystyle p = \frac{3}{4}.
- Diverges: \displaystyle 2^{-\ln n} = e^{-(\ln 2)(\ln n)} = \frac{1}{n^{\ln 2}}, and \ln 2 \approx 0.693 < 1.
- Converges: \displaystyle 3^{-\ln n} = \frac{1}{n^{\ln 3}}, and \ln 3 \approx 1.099 > 1.
- Converges: \displaystyle 5 \cdot \frac{1}{n^{6/5}}, \displaystyle p = \frac{6}{5}; a constant factor does not change convergence.
- Exactly when 3 - 2p > 1, that is p < 1.
- S_5 \approx 1.080352 and \displaystyle \frac{1}{3 \cdot 6^3} \le R_5 \le \frac{1}{3 \cdot 5^3}, that is \displaystyle \frac{1}{648} \le R_5 \le \frac{1}{375}. So 1.081895 \le S \le 1.083019. (The sum is \displaystyle \frac{\pi^4}{90} \approx 1.082323.)
- \displaystyle R_N \le \int_N^{\infty} x^{-3/2}\,dx = \frac{2}{\sqrt{N}} \le 0.01 when \displaystyle \sqrt{N} \ge 200: N = 40{,}000 terms.
- About 1.203287, with error at most 0.005625. \displaystyle S_4 = 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{64} \approx 1.177662 and \displaystyle \frac{1}{2 \cdot 5^2} \le R_4 \le \frac{1}{2 \cdot 4^2}, so S lies between 1.197662 and 1.208912. (The sum is 1.202057 to six decimals.)
- The fraction is \displaystyle \frac{R_{10}}{S} for \displaystyle \sum \frac{1}{n^3}. \displaystyle R_{10} \le \frac{1}{2 \cdot 10^2} = 0.005, and S > 1 (the first term alone), so the fraction is below 0.005 = 0.5\%. (In fact about 0.38\%.)
- Between 651 and 676. The bounds read \displaystyle 2\sqrt{N+1} - 2 \le S_N \le 2\sqrt{N} - 1. S_N > 50 needs \displaystyle 2\sqrt{N} - 1 > 50, N > 650.25; and \displaystyle 2\sqrt{N+1} - 2 > 50 once N + 1 > 676. (Directly, the first N is 662.)
- The bounds on the sum with m = 1, using \displaystyle \int_1^{\infty} x^{-p}\,dx = \frac{1}{p-1}, give \displaystyle \frac{1}{p-1} \le S(p) \le 1 + \frac{1}{p-1}, and \displaystyle 1 + \frac{1}{p-1} = \frac{p}{p-1}. Multiplying by p - 1 > 0, 1 \le (p-1)S(p) \le p, and squeezing as p \to 1^+ gives the limit 1.
- The bounds on partial sums put S_N above \displaystyle \frac{(N+1)^{1-p} - 1}{1-p} and below \displaystyle 1 + \frac{N^{1-p} - 1}{1-p}. Divide by N^{1-p}, which tends to \infty: since \displaystyle \left(\frac{N+1}{N}\right)^{1-p} \to 1, both sides tend to \displaystyle \frac{1}{1-p}, and the squeeze theorem gives the limit.