TAOCP 1.2.7: Harmonic Numbers
Section 1.2.7 exercises: 25/25 solved.
Section 1.2.7. Harmonic Numbers
Exercises from TAOCP Volume 1 Section 1.2.7: 25/25 solved.
| # | Rating | Category | Status | Time |
|---|---|---|---|---|
| 1 | [01] | simple | solved | - |
| 2 | [13] | simple | solved | - |
| 3 | [M21] | math-medium | solved | - |
| 4 | [10] | simple | solved | - |
| 5 | [15] | simple | solved | - |
| 6 | [M15] | math-simple | solved | - |
| 7 | [M21] | math-medium | solved | - |
| 8 | [HM18] | hm-medium | verified | 1m23s |
| 9 | [M18] | math-medium | solved | - |
| 10 | [M20] | math-medium | solved | - |
| 11 | [M21] | math-medium | verified | 1m14s |
| 12 | [M10] | math-simple | solved | - |
| 13 | [M22] | math-medium | solved | - |
| 14 | [M22] | math-medium | solved | - |
| 15 | [M23] | math-medium | solved | - |
| 16 | [18] | medium | solved | - |
| 17 | [M24] | math-medium | solved | - |
| 18 | [M33] | math-hard | solved | - |
| 19 | [M30] | math-hard | solved | - |
| 20 | [HM22] | hm-medium | solved | 4m12s |
| 21 | [M24] | math-medium | solved | - |
| 22 | [M28] | math-hard | solved | - |
| 23 | [HM20] | hm-medium | solved | - |
| 24 | [HM21] | hm-medium | verified | 2m38s |
| 25 | [M21] | math-medium | solved | - |
TAOCP 1.2.7 Exercise 1
By definition, H_n = \sum_{k=1}^{n}\frac1k.
TAOCP 1.2.7 Exercise 2
The argument in the text groups the terms of $H_{2^m}$ as follows: H_{2^m} = 1 + \frac12 + \left(\frac13 + \frac14\right) + \left(\frac15 + \cdots + \frac18\right)
TAOCP 1.2.7 Exercise 3
Let H_n^{(r)}=\sum_{k=1}^{n}\frac{1}{k^r}, \qquad r>1.
TAOCP 1.2.7 Exercise 4
Statement 1 is false, since $H_1=1>\ln 1=0$.
TAOCP 1.2.7 Exercise 5
By the asymptotic expansion for harmonic numbers in Appendix A, H_n = \ln n + \gamma + \frac{1}{2n} -\frac{1}{12n^2} +\frac{1}{120n^4} -\frac{1}{252n^6}
TAOCP 1.2.7 Exercise 6
We wish to prove that H_n = \frac{\left[{\,n+1 \atop 2\,}\right]}{n!
TAOCP 1.2.7 Exercise 7
Let T(m,n) = H_m + H_n - H_{mn}, where $m,n > 0$ and $H_k = \sum_{i=1}^{k} \frac{1}{i}$ is the $k$-th harmonic number.
TAOCP 1.2.7 Exercise 8
Equation (8) gives \sum_{k=1}^{n}H_k=(n+1)H_n-n.
TAOCP 1.2.7 Exercise 9
We are asked to evaluate \sum_{k=1}^{n} \binom{n}{k}(-1)^k H_k.
TAOCP 1.2.7 Exercise 10
Let S=\sum_{1 \le k < n}(a_{k+1}-a_k)b_k.
TAOCP 1.2.7 Exercise 11
Let S=\sum_{1<k\le n}\frac{1}{k(k-1)}H_k.
TAOCP 1.2.7 Exercise 12
By the definition of generalized harmonic numbers, H_{\infty}^{(1000)}=\sum_{k=1}^{\infty}\frac1{k^{1000}}=\zeta(1000).
TAOCP 1.2.7 Exercise 13
We prove the identity \sum_{k=1}^{n}\frac{x^k}{k} = H_n + \sum_{k=1}^{n}\binom{n}{k}\frac{(x-1)^k}{k}.
TAOCP 1.2.7 Exercise 14
Write S_n=\sum_{k=1}^{n}\frac{H_k}{k}.
TAOCP 1.2.7 Exercise 15
Let S_n=\sum_{k=1}^{n}H_k^2.
TAOCP 1.2.7 Exercise 16
Let S_n = 1 + \frac{1}{3} + \cdots + \frac{1}{2n-1} = \sum_{k=1}^{n}\frac{1}{2k-1}.
TAOCP 1.2.7 Exercise 17
Since $p$ is an odd prime, the integers 1,2,\ldots,p-1 can be paired as
TAOCP 1.2.7 Exercise 18
Let S_n = 1 + \frac13 + \frac15 + \cdots + \frac{1}{2n-1}.
TAOCP 1.2.7 Exercise 19
Let H_n=\sum_{k=1}^{n}\frac{1}{k}, \qquad H_0=0.
TAOCP 1.2.7 Exercise 20
Let b_k=a_kx_0^k, \qquad f(x_0)=\sum_{k\ge0}b_k,
TAOCP 1.2.7 Exercise 21
Let S_n=\sum_{k=1}^{n}\frac{H_k}{n+1-k}.
TAOCP 1.2.7 Exercise 22
Let S_n=\sum_{k=0}^{n} H_kH_{n-k}.
TAOCP 1.2.7 Exercise 23
For positive integers $n$, H_n=\sum_{k=1}^n \frac1k.