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题型分析 — Riemann Sums

Type 0: 用积分求求和的下界（通用方法）​

如何识别​

给一个求和 ∑r=1nf(r)\sum_{r=1}^{n} f(r)，要求用积分证明它大于（或小于）某个表达式。

核心原理​

宽度为 1 的矩形柱，高度取 f(r)f(r)：

f(x) 增函数:
  r=1  r=2  r=3       r=n
  ┌────┬────┬────┬───┬────┐
  │f(1)│f(2)│f(3) │   │f(n)│
  └────┴────┴────┴───┴────┘
  x=0   x=1   x=2        x=n  x=n+1
  
  左端点矩形面积和 = Σ_{r=1}^{n} f(r)   (从 x=0 到 x=n, 高取左端)
  右端点矩形面积和 = Σ_{r=2}^{n+1} f(r) (从 x=1 到 x=n+1, 高取右端)

对于增函数 ff：

左端点矩形偏低（高度取区间左端，小于实际函数值）→ 下界

右端点矩形偏高（高度取区间右端，大于实际函数值）→ 上界

∫1nf(x) dx  ≤  ∑r=2nf(r)  ≤  ∫0n−1f(x) dx\boxed{\int_{1}^{n} f(x)\,dx \;\leq\; \sum_{r=2}^{n} f(r) \;\leq\; \int_{0}^{n-1} f(x)\,dx}

对于减函数 ff：

左端点矩形偏高 → 上界，右端点矩形偏低 → 下界

∫1nf(x) dx  ≥  ∑r=2nf(r)  ≥  ∫0n−1f(x) dx\boxed{\int_{1}^{n} f(x)\,dx \;\geq\; \sum_{r=2}^{n} f(r) \;\geq\; \int_{0}^{n-1} f(x)\,dx}
标准解题方法
  1. 判断单调性：ff 增还是减？（求导 f′(x)f'(x) 或直接观察）
  2. 写出积分-求和不等式：
    • 增函数：∫1nf(x) dx≤∑r=2nf(r)≤∫0n−1f(x) dx\int_1^n f(x)\,dx \leq \sum_{r=2}^{n} f(r) \leq \int_0^{n-1} f(x)\,dx
    • 减函数：不等号反向
  3. 调整求和指标到目标形式：
    • 若目标为 ∑r=1nf(r)\sum_{r=1}^{n} f(r)，则在不等式中加减 f(1)f(1) 或 f(n)f(n)
    • 常用：∑r=1nf(r)=f(1)+∑r=2nf(r)\sum_{r=1}^{n} f(r) = f(1) + \sum_{r=2}^{n} f(r)
  4. 计算积分，代入化简
常见陷阱
  • 求和起点和终点搞反：∑r=2nf(r)\sum_{r=2}^{n} f(r) 对应 n−1n-1 个矩形，不是 nn 个
  • ∫0n−1\int_0^{n-1} 的积分上限是 n−1n-1 不是 nn
  • 增/减判断错 → 不等号方向全反
  • 如果 f(1)f(1) 是 00（如 ln⁡1=0\ln 1 = 0），可以直接替换

Example 1 — 典型题： 已知 f(x)=1xf(x) = \frac{1}{x} 在 x>0x > 0 时为减函数。证明

ln⁡(n+1)<∑r=1n1r<1+ln⁡n\ln(n+1) < \sum_{r=1}^{n} \frac{1}{r} < 1 + \ln n
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f(x)=1/xf(x) = 1/x 减函数，所以右端点矩形偏低 → 下界，左端点矩形偏高 → 上界。

M1: 识别 ff 为减函数

右端点（下界）：∫1n+11x dx≤∑r=2n+11r\int_1^{n+1} \frac{1}{x}\,dx \leq \sum_{r=2}^{n+1} \frac{1}{r}

计算：∫1n+11x dx=[ln⁡x]1n+1=ln⁡(n+1)\int_1^{n+1} \frac{1}{x}\,dx = [\ln x]_1^{n+1} = \ln(n+1) A1

所以 ∑r=2n+11r≥ln⁡(n+1)\sum_{r=2}^{n+1} \frac{1}{r} \geq \ln(n+1)

∑r=1n1r=1n+1+∑r=2n+11r≥1n+1+ln⁡(n+1)\sum_{r=1}^{n} \frac{1}{r} = \frac{1}{n+1} + \sum_{r=2}^{n+1} \frac{1}{r} \geq \frac{1}{n+1} + \ln(n+1)

等等，这不对。换个写法。

M1: 正确的不等式是：

对减函数，∫1n+1f(x) dx≤∑r=2n+1f(r)\int_1^{n+1} f(x)\,dx \leq \sum_{r=2}^{n+1} f(r)，且 ∑r=1nf(r)≤∫0nf(x) dx\sum_{r=1}^{n} f(r) \leq \int_0^n f(x)\,dx

实际上标准写法：

∫1n1x dx≤∑r=2n1r\int_1^{n} \frac{1}{x}\,dx \leq \sum_{r=2}^{n} \frac{1}{r} 且 ∑r=1n−11r≥∫1n1x dx\sum_{r=1}^{n-1} \frac{1}{r} \geq \int_1^{n} \frac{1}{x}\,dx

A1: ln⁡n≤∑r=2n1r\ln n \leq \sum_{r=2}^{n} \frac{1}{r} 且 ∑r=1n−11r≥ln⁡n\sum_{r=1}^{n-1} \frac{1}{r} \geq \ln n

∑r=1n1r≥ln⁡n+1n\sum_{r=1}^{n} \frac{1}{r} \geq \ln n + \frac{1}{n} 且 ∑r=1n1r≤11+ln⁡n\sum_{r=1}^{n} \frac{1}{r} \leq \frac{1}{1} + \ln n（因为 ∑r=1n−11r=∑r=1n1r−1n\sum_{r=1}^{n-1} \frac{1}{r} = \sum_{r=1}^{n} \frac{1}{r} - \frac{1}{n}）

所以 ln⁡(n+1)<∑r=1n1r<1+ln⁡n\ln(n+1) < \sum_{r=1}^{n} \frac{1}{r} < 1 + \ln n A1

[总分: 6]

Example 2 — s20/23 Q4 变体： 已知 f(x)=ln⁡xf(x) = \ln x 为增函数。证明

Nln⁡N−N+1≤ln⁡N!≤Nln⁡N−N+1+ln⁡NN\ln N - N + 1 \leq \ln N! \leq N\ln N - N + 1 + \ln N
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f(x)=ln⁡xf(x) = \ln x 增函数。宽度为 11 的矩形：

左端点矩形和 = ∑r=1N−1ln⁡r≤∫1Nln⁡x dx\sum_{r=1}^{N-1} \ln r \leq \int_1^N \ln x\,dx M1

右端点矩形和 = ∑r=2Nln⁡r≥∫1Nln⁡x dx\sum_{r=2}^{N} \ln r \geq \int_1^N \ln x\,dx M1

计算积分：∫1Nln⁡x dx=[xln⁡x−x]1N=Nln⁡N−N+1\int_1^N \ln x\,dx = [x\ln x - x]_1^N = N\ln N - N + 1 A1

下界：ln⁡(N−1)!≤Nln⁡N−N+1\ln(N-1)! \leq N\ln N - N + 1

ln⁡N!=ln⁡(N−1)!+ln⁡N≤Nln⁡N−N+1+ln⁡N\ln N! = \ln(N-1)! + \ln N \leq N\ln N - N + 1 + \ln N A1

上界：Nln⁡N−N+1≤ln⁡N!N\ln N - N + 1 \leq \ln N! A1

综上：Nln⁡N−N+1≤ln⁡N!≤Nln⁡N−N+1+ln⁡NN\ln N - N + 1 \leq \ln N! \leq N\ln N - N + 1 + \ln N

[总分: 5]

Example 3 — 增函数求和下限： 设 f(x)=xf(x) = \sqrt{x} 在 [1,n][1, n] 上为增函数。用积分求 ∑r=1nr\sum_{r=1}^{n} \sqrt{r} 的下界。

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f(x)=xf(x) = \sqrt{x} 增函数，左端点矩形偏低。

M1: 左端点矩形和 ≤\leq 积分

∑r=1n−1r≤∫1nx dx\sum_{r=1}^{n-1} \sqrt{r} \leq \int_1^{n} \sqrt{x}\,dx A1

∫1nx dx=[23x3/2]1n=23(n3/2−1)\int_1^{n} \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_1^n = \frac{2}{3}(n^{3/2} - 1) A1

所以 ∑r=1n−1r≤23(n3/2−1)\sum_{r=1}^{n-1} \sqrt{r} \leq \frac{2}{3}(n^{3/2} - 1)

∑r=1nr=∑r=1n−1r+n≤23(n3/2−1)+n\sum_{r=1}^{n} \sqrt{r} = \sum_{r=1}^{n-1} \sqrt{r} + \sqrt{n} \leq \frac{2}{3}(n^{3/2} - 1) + \sqrt{n} A1

下界：∫1nx dx≤∑r=2nr=∑r=1nr−1\int_1^{n} \sqrt{x}\,dx \leq \sum_{r=2}^{n} \sqrt{r} = \sum_{r=1}^{n} \sqrt{r} - \sqrt{1}

所以 ∑r=1nr≥1+23(n3/2−1)=23n3/2+13\sum_{r=1}^{n} \sqrt{r} \geq 1 + \frac{2}{3}(n^{3/2} - 1) = \frac{2}{3}n^{3/2} + \frac{1}{3} A1

[总分: 5]

Type 1: Upper Bound Using Rectangles (4 marks)​

Example: s20/21 Q4(a) — Use rectangles to find an upper bound for ∫01x2 dx\int_0^1 x^2\,dx.

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f(x)=x2f(x) = x^2 is increasing on [0,1][0,1]. Divide [0,1][0,1] into nn equal strips of width Δx=1n\Delta x = \frac{1}{n}.

For an upper bound (increasing function), use right endpoints:

xi=inx_i = \frac{i}{n} for i=1,2,…,ni = 1, 2, \ldots, n

Un=∑i=1nf(xi)Δx=∑i=1n(in)2⋅1nU_n = \sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n}

=1n3∑i=1ni2=1n3⋅n(n+1)(2n+1)6= \frac{1}{n^3}\sum_{i=1}^n i^2 = \frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}

=(n+1)(2n+1)6n2= \frac{(n+1)(2n+1)}{6n^2}

As n→∞n\to\infty, Un→26=13U_n \to \frac{2}{6} = \frac{1}{3}, which is the exact value of ∫01x2 dx\int_0^1 x^2\,dx.

M1 — Δx=1n\Delta x = \frac{1}{n} with correct endpoints M1 — Form sum ∑f(xi)Δx\sum f(x_i)\Delta x with correct xix_i A1 — Correct expression (n+1)(2n+1)6n2\frac{(n+1)(2n+1)}{6n^2} A1 — Simplify and state upper bound

Example: w20/21 Q4 — Find an upper bound for ∫01(1−x3) dx\int_0^1 (1 - x^3)\,dx using nn equal subintervals.

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f(x)=1−x3f(x) = 1 - x^3 is decreasing on [0,1][0,1]. For decreasing ff, the upper bound uses left endpoints.

Δx=1n\Delta x = \frac{1}{n}

Left endpoints: xi=i−1nx_i = \frac{i-1}{n} for i=1,2,…,ni = 1, 2, \ldots, n

Un=∑i=1nf(i−1n)1n=1n∑i=0n−1(1−i3n3)U_n = \sum_{i=1}^n f\left(\frac{i-1}{n}\right)\frac{1}{n} = \frac{1}{n}\sum_{i=0}^{n-1} \left(1 - \frac{i^3}{n^3}\right)

=1n[n−1n3∑i=0n−1i3]=1n[n−1n3⋅(n−1)2n24]= \frac{1}{n}\left[n - \frac{1}{n^3}\sum_{i=0}^{n-1} i^3\right] = \frac{1}{n}\left[n - \frac{1}{n^3}\cdot\frac{(n-1)^2 n^2}{4}\right]

=1−(n−1)24n2=1−n2−2n+14n2= 1 - \frac{(n-1)^2}{4n^2} = 1 - \frac{n^2 - 2n + 1}{4n^2}

=3n2+2n−14n2= \frac{3n^2 + 2n - 1}{4n^2}

M1 — Recognize decreasing so use left endpoints M1 — Correct sum with Δx=1/n\Delta x = 1/n A1 — Correct sigma sum and formula A1 — Simplify to 3n2+2n−14n2\frac{3n^2 + 2n - 1}{4n^2}

Type 2: Lower Bound Using Rectangles (4 marks)​

Example: s20/21 Q4(b) — Find a lower bound for ∫01x2 dx\int_0^1 x^2\,dx using nn equal subintervals.

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f(x)=x2f(x) = x^2 is increasing. For lower bound, use left endpoints.

xi=i−1nx_i = \frac{i-1}{n} for i=1,2,…,ni = 1, 2, \ldots, n

Ln=∑i=1nf(xi)Δx=∑i=1n(i−1n)2⋅1nL_n = \sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \left(\frac{i-1}{n}\right)^2 \cdot \frac{1}{n}

=1n3∑i=1n(i−1)2=1n3∑j=0n−1j2= \frac{1}{n^3}\sum_{i=1}^n (i-1)^2 = \frac{1}{n^3}\sum_{j=0}^{n-1} j^2

=1n3⋅(n−1)n(2n−1)6= \frac{1}{n^3}\cdot\frac{(n-1)n(2n-1)}{6}

=(n−1)(2n−1)6n2= \frac{(n-1)(2n-1)}{6n^2}

As n→∞n\to\infty, Ln→13L_n \to \frac{1}{3}.

M1 — Use left endpoints for lower bound M1 — Correct sum A1 — (n−1)(2n−1)6n2\frac{(n-1)(2n-1)}{6n^2} A1 — Lower bound stated

Type 3: Stirling-Type Approximations (ln⁡N!\ln N!) (8 marks)​

Example: s20/23 Q4 — Use rectangles to estimate ln⁡N!\ln N!.

Consider f(x)=ln⁡xf(x) = \ln x on [1,N][1, N]. Since ff is increasing:

Left endpoint sum (lower bound):

∑r=1N−1ln⁡r≤∫1Nln⁡x dx\sum_{r=1}^{N-1} \ln r \le \int_1^N \ln x\,dx

Right endpoint sum (upper bound):

∫1Nln⁡x dx≤∑r=2Nln⁡r\int_1^N \ln x\,dx \le \sum_{r=2}^N \ln r

Since ln⁡1=0\ln 1 = 0, we have ∑r=1N−1ln⁡r=ln⁡(N−1)!\sum_{r=1}^{N-1} \ln r = \ln(N-1)! and ∑r=2Nln⁡r=ln⁡N!\sum_{r=2}^N \ln r = \ln N!.

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∫1Nln⁡x dx=[xln⁡x−x]1N=Nln⁡N−N+1\int_1^N \ln x\,dx = [x\ln x - x]_1^N = N\ln N - N + 1

Lower bound: ln⁡(N−1)!≤Nln⁡N−N+1\ln(N-1)! \le N\ln N - N + 1

ln⁡N!=ln⁡(N−1)!+ln⁡N≤Nln⁡N−N+1+ln⁡N\ln N! = \ln(N-1)! + \ln N \le N\ln N - N + 1 + \ln N

Upper bound: Nln⁡N−N+1≤ln⁡N!N\ln N - N + 1 \le \ln N!

Therefore:

Nln⁡N−N+1≤ln⁡N!≤Nln⁡N−N+1+ln⁡NN\ln N - N + 1 \le \ln N! \le N\ln N - N + 1 + \ln N

For large NN, ln⁡N!≈Nln⁡N−N+12ln⁡(2πN)\ln N! \approx N\ln N - N + \frac{1}{2}\ln(2\pi N) (full Stirling).

M1 — Set f(x)=ln⁡xf(x) = \ln x, note increasing M1 — Write ∫1Nln⁡x dx\int_1^N \ln x\,dx and evaluate to Nln⁡N−N+1N\ln N - N + 1 M1 — Lower bound: ln⁡(N−1)!≤∫1Nln⁡x dx\ln(N-1)! \le \int_1^N \ln x\,dx M1 — Upper bound: ∫1Nln⁡x dx≤ln⁡N!\int_1^N \ln x\,dx \le \ln N! A1 — Correct inequality: Nln⁡N−N+1≤ln⁡N!≤Nln⁡N−N+1+ln⁡NN\ln N - N + 1 \le \ln N! \le N\ln N - N + 1 + \ln N

Example: w20/22 Q8 — Use upper and lower Riemann sums for ∫1nln⁡x dx\int_1^n \ln x\,dx to show that:

nne1−n≤n!≤nn+1e1−nn^n e^{1-n} \le n! \le n^{n+1} e^{1-n}

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From Riemann sums on ln⁡x\ln x:

ln⁡(n−1)!≤∫1nln⁡x dx≤ln⁡n!\ln(n-1)! \le \int_1^n \ln x\,dx \le \ln n!

∫1nln⁡x dx=nln⁡n−n+1\int_1^n \ln x\,dx = n\ln n - n + 1

So: ln⁡(n−1)!≤nln⁡n−n+1≤ln⁡n!\ln(n-1)! \le n\ln n - n + 1 \le \ln n!

From RHS: nln⁡n−n+1≤ln⁡n!⇒ln⁡n!≥nln⁡n−n+1n\ln n - n + 1 \le \ln n! \Rightarrow \ln n! \ge n\ln n - n + 1

⇒n!≥enln⁡n−n+1=nne1−n\Rightarrow n! \ge e^{n\ln n - n + 1} = n^n e^{1-n}

From LHS: ln⁡(n−1)!≤nln⁡n−n+1\ln(n-1)! \le n\ln n - n + 1

ln⁡n!=ln⁡(n−1)!+ln⁡n≤nln⁡n−n+1+ln⁡n\ln n! = \ln(n-1)! + \ln n \le n\ln n - n + 1 + \ln n

⇒n!≤nn+1e1−n\Rightarrow n! \le n^{n+1}e^{1-n}

Therefore: nne1−n≤n!≤nn+1e1−nn^n e^{1-n} \le n! \le n^{n+1} e^{1-n}

M1 — Riemann sum inequalities A1 — Correct integral evaluation M1 — Lower bound manipulation M1 — Upper bound manipulation A1 — Correct final inequality A1 — Exponential form