Exercises from Section 1.2.7

Tord M. Johnson

April 27, 2015

1. [01] What are H0, H1, and H2?

By definition, we have

H0 = 1k0 1 k = 0,
H1 = 1k1 1 k = 1 1 = 1,

and

H2 = 1k2 1 k = 1 1 + 1 2 = 3 2.

2. [13] Show that the simple argument used in the text to prove that H2m 1 + m2 can be slightly modified to prove that H2m 1 + m.

We can show that the simple argument used in the text to prove that H2m 1 + m2 may be slightly modified to prove that H2m 1 + m, by noting that for each term, 1(2m + k) 12m, as shown in the proof by induction below.

Proposition. H2m m + 1.

Proof. Let m be an arbitrary integer such that m 0. We must show that H2m m + 1. In the case that m = 0,

H20 = H1 = 1 0 + 1.

Then, assuming

H2m m + 1

we must show that

H2m+1 m + 2.

But

H2m+1 = 1k2m+1 1 k = 1k2m1 k + 2m+1k2m+1 1 k = H2m + 2m+1k2m+1 1 k = H2m + 1k2m 1 2m + k H2m + 1k2m 1 2m = H2m + 2m 2m = H2m + 1 m + 1 + 1 = m + 2

as we needed to show. □

3. [M21] Generalize the argument used in the previous exercise to show that, for r > 1, the sum Hn(r) remains bounded for all n. Find an upper bound.

Proposition. Hn(r) 2r1 2r11 for r > 1.

Proof. Let n be an arbitrary nonnegative integer and r an arbitrary real such that r > 1. We must show that

Hn(r) 2r1 2r1 1.

First note that for arbitrary m 1, we may show that

1k2m1 1 kr 0k<m 2k 2kr.

If m = 1,

1k211 1 kr = 1k0 1 kr = 0 1 = 20 2(0)r = 0k<1 2k 2kr.

Then assuming

1k2m1 1 kr 0k<m 2k 2kr.

we must show that

1k2m 1 kr 0k<m+1 2k 2kr.

But

1k2m 1 kr = 1k2m1 1 kr + 2m1+1k2m 1 kr 0k<m 2k 2kr + 2m1+1k2m 1 kr = 0k<m 2k 2kr + 1k2m1 1 (2m1 + k)r 0k<m 2k 2kr + 1k2m1 1 (2m1)r = 0k<m 2k 2kr + 2m1 (2m1)r = 0k<m 2k 2kr + 2m1 2(m1)r 0k<m 2k 2kr + 2m 2mr = 0k<m+1 2k 2kr

and hence the noted inequality. We now continue with the main proof.

Since 2r1 > 1, we have both in the case that n = 0 that

H0(r) = 1k0 1 kr = 0 2r1 2r1 1

and in the case that n = 1 = 2m1 for m = 1 that

H1(r) = 1k1 1 kr = 1 2r1 2r1 1.

Then, for arbitrary m 1, and since 2mr+m+r1 = 2r1 2(r1)m < 1 < 2r1,

H2m1(r) = 1k2m1 1 kr 0k<m 2k 2kr = 0k<m 1 2(r1)k = 0km12(r+1)k = 2(r+1)0 2(r+1)m 1 2r+1 = 1 2(r+1)m 1 2r+1 = (2m(r1) 1)2m(r1) (2r1 1)2r1 = 2r1(2m(r1) 1) 2m(r1)(2r1 1) = 2m(r1)+(r1) 2r1 2m(r1)+(r1) 2m(r1) = 2m(r1)(2m(r1)+(r1) 2r1) 2r1 1 = 2m(r1)2m(r1)+(r1) 2m(r1)2r1 2r1 1 = 2r1 2m(r1)+r1 2r1 1 = 2r1 2mr+m+r1 2r1 1 2r1 2r1 1

as we needed to show. □

4. [10] Decide which of the following statements are true for all positive integers n: (a) Hn < ln n. (b) Hn > ln n. (c) Hn > ln n + γ.

In summary, (a) is false, while (b) and (c) are true, the justification for each enumerated below.

a)
Hn < ln n is not true for all positive integers n, as may be seen by considering n = 1, in which case, H1 = 1 0 = ln 1.
b)
Hn > ln n is true for all positive integers n, as may be deduced from Eq. (3), since γ + 1 2n 1 12n2 + 1 120n4 𝜖 > 0.
c)
Hn > ln n + γ is true for all positive integers n, as may also be deduced from Eq. (3), since 1 2n 1 12n2 + 1 120n4 𝜖 > 0.

5. [15] Give the value of H10000 to 15 decimal places, using the tables in Appendix A.

From Eq. (3) we know

H10000 = ln 10000 + γ + 1 2(10000) 1 12(10000)2 + 1 120(10000)4 𝜖

for 0 < 𝜖 < 1 252(10000)6 . Letting 𝜖 = 1 120(10000)4 𝜖 > 0, since

𝜖 < 1 120(10000)4 = 1 1.2 × 1018 < 1 1018,

we may ignore 𝜖 in order to approximate H10000 to only 15 decimal places as

H10000 ln 10000 + γ + 1 2(10,000) 1 12(10,000)2 = 4ln 10 + γ + 59999 1200000000.

Given

ln 10 = 2.3025850929940456+ γ = 0.5772156649015328+ 59999 1200000000 = 0.0000499991666666+

we may compute the sum as

2.30258 50929 94045 6
2.30258 50929 94045 6
2.30258 50929 94045 6
2.30258 50929 94045 6
0.57721 56649 01532 8
+ 0.00004 99991 66666 6


9.78760 60360 44381 8

That is,

H10000 9.787606036044382

6. [M15] Prove that the harmonic numbers are directly related to Stirling’s numbers, which were introduced in the previous section; in fact,

Hn = n + 1 2 n!.

Proposition. Hn = n+1 2 n!.

Proof. Let n be an arbitrary nonnegative integer. We must show that

Hn = n + 1 2 n!.

In the case that n = 0,

H0 = 1k0 1 k = 0 = 1 2 = 0 + 1 2 0!.

Then, assuming

Hn = n + 1 2 n!

we must show that

Hn+1 = n + 2 2 (n + 1)!.

But

Hn+1 = Hn + 1 n + 1 = n + 1 2 n! + 1 n + 1 = (n + 1)n + 1 2 + n!(n + 1)! = (n + 1)n + 1 2 + n + 1 1 (n + 1)! from Eq. (50) = (n + 1)n + 1 2 + n + 1 2 1 (n + 1)! = n + 2 2 (n + 1)! from Eq. (46)

as we needed to show. □

7. [M21] Let T(m,n) = Hm + Hn Hmn. (a) Show that when m or n increases, T(m,n) never increases (assuming that m and n are positive). (b) Compute the minimum and maximum values of T(m,n) for m,n > 0.

We may provide a proof and determine bounds.

a)
We may show that T(m,n) never increases.

Proposition. T(m + 1,n) T(m,n) for m,n positive integers.

Proof. Define T(m,n) as

T(m,n) = Hm + Hn Hmn

and let m and n be arbitrary positive integers. We must show that

T(m + 1,n) T(m,n) 0.

But

T(m + 1,n) T(m,n) = Hm+1 + Hn H(m+1)n Hm + Hn Hmn = Hm+1 + Hn H(m+1)n Hm Hn + Hmn = Hm+1 H(m+1)n Hm + Hmn = 1 m + 1 mn+1kmn+n1 k 1 m + 1 mn+1kmn+n 1 mn + n = 1 m + 1 n mn + n = 1 m + 1 1 m + 1 = 0

as we needed to show. □

b)
We may determine both the lower and upper bounds of T(m,n), for m,n positive integers. Since T(m,n) never increases, we know that the lower bound corresponds to the limit as m , and from Eq. (3),
lim mT(m,n) = lim mHm + Hn Hmn = lim mHm ln m = γ.

Similarly, since T(m,n) never increases, we know that the upper bound corresponds to m = n = 1, and

T(1,1) = H1 + H1 H1 = H1 = 1.

______________________________________________________________________________________________________________________________

[AMM 70 (1963), 575–577]

8. [HM18] Compare Eq. (8) with k=1n ln k; estimate the difference as a function of n.

Given Eq. (8)

1knHk = (n + 1)Hn n

we may estimate the difference with 1kn ln k. First, we note from Eq. (3) that

1knHk = (n + 1)Hn n (n + 1) ln n + γ + 12n n = (n + 1)ln n + (n + 1)γ + (n + 1)2n n = (n + 1)ln n n + (n + 1)γ + (n + 1)2n (n + 1)ln n n + (n + 1)γ + 12 = (n + 1)ln n n + nγ + γ + 12 = (n + 1)ln n n(1 γ) + (γ + 12).

Second, we note from Stirling’s approximation that

1kn ln k = ln n! ln 2πn n en = ln 2π + 1 2ln n + nln n nln e = ln 2π + n + 1 2ln n n = n + 1 2ln n n + ln 2π.

And so,

1knHk 1kn ln k (n + 1)ln n n(1 γ) + γ + 1 2 n + 1 2ln n n + ln 2π = (n + 1)ln n + n + γn + γ + 1 2 n + 1 2ln n + n ln 2π = γn + n + 1 n 1 2ln n + γ + 1 2 ln 2π = γn + 1 2ln n + γ + 1 2 ln 2π γn + 1 2ln n + .158.

9. [M18] Theorem A applies only when x > 0; what is the value of the sum considered when x = 1?

We make a proposition and offer proof in the case that x = 1.

Proposition. 1knn k (1)kH k = 1 n.

Proof. Let n be an arbitrary positive integer. We must show that

1knn k (1)kH k = 1 n.

If n = 1,

1k1 1 k(1)kH k = 1 1(1)1H 1 = 1 1.

Then, assuming

1knn k (1)kH k = 1 n,

we must show that

1kn+1n + 1 k (1)kH k = 1 n + 1.

But

1kn+1n + 1 k (1)kH k = 1kn+1 n k + n k 1(1)kH k = 1kn+1n k (1)kH k + 1kn+1 n k 1(1)kH k = 1knn k (1)kH k + 1kn+1 n k 1(1)kH k = 1 n + 1kn+1 n k 1(1)kH k = 1 n 1kn+1 n k 1(1)k1H k = 1 n 1kn+1 n k 1(1)k1 H k1 + 1 k = 1 n 1kn+1 n k 1(1)k1H k1 1kn+1 n k 1(1)k1 1 k = 1 n 0knn k (1)kH k 1kn+1 n k 1(1)k1 1 k = 1 n 1knn k (1)kH k 1kn+1 n k 1(1)k1 1 k = 1 n + 1 n 1kn+1 n k 1(1)k1 1 k = 1kn+1 n k 1(1)k1 1 k = 1kn+1 1 n + 1 n + 1 k (1)k1 from Eq. 1.2.6-(7) = 1 n + 1 1kn+1n + 1 k (1)k = 1 n + 1 n + 1 0 (1)0 + 0kn+1n + 1 k (1)k = 1 n + 1 1 + (1 1)n+1 = 1 n + 1

as we needed to show. □

10. [M20] (Summation by parts.) We have used special cases of the general method of summation by parts in exercise 1.2.4-42 and in the derivation of Eq. (9). Prove the general formula

1k<n(ak+1 ak)bk = anbn a1b1 1k<nak+1(bk+1 bk).

Proposition. 1k<n(ak+1 ak)bk = anbn a1b1 1k<nak+1(bk+1 bk).

Proof. Let n be an arbitrary positive integer. We must show that

1k<n(ak+1 ak)bk = anbn a1b1 1k<nak+1(bk+1 bk).

But

1k<n(ak+1 ak)bk = 1k<nak+1bk 1k<nakbk = 1k<nak+1bk 0k<n1ak+1bk+1 = 1k<nak+1bk a1b1 + 1k<nak+1bk+1 anbn = anbn a1b1 + 1k<nak+1bk 1k<nak+1bk+1 = anbn a1b1 1k<nak+1bk+1 1k<nak+1bk = anbn a1b1 1k<nak+1(bk+1 bk)

as we needed to show. □

11. [M21] Using summation by parts, evaluate

1<kn 1 k(k 1)Hk.

The sum may be evaluated using summation by parts as

1<kn 1 k(k 1)Hk = 1<knk (k 1) k(k 1) Hk = 1<kn 1 k 1 1 kHk = 1<kn 1 (k + 1) 1 1 k 1Hk = 1k<n 1 k + 1 1 kHk+1 = 1 nHn+1 1 1H1+1 1k<n 1 k + 1 H(k+1)+1 Hk+1 = 1 n Hn + 1 n + 1 + 1 + 1 2 + 1k<n 1 k + 1 H(k+1)+1 Hk+1 = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + 1k<n 1 k + 1 H(k+1)+1 Hk+1 = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + 1k<n 1 k + 1 1 k + 2 = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + 1k<n 1 (k + 1)(k + 2) = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + 1k<n(k + 2) (k + 1) (k + 1)(k + 2) = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + 1k<n 1 k + 1 1 k + 2 = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + 1k<n 1 k + 1 1k<n 1 k + 2 = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + 2kn1 k 3kn+1 1 k = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + 1kn1 k 1 1kn1 k 1 1 2 + 1 n + 1 = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + Hn 1 Hn 1 1 2 + 1 n + 1 = 1 nHn 1 n 1 n + 1 + 1 + 1 2 + Hn 1 Hn + 1 + 1 2 1 n + 1 = 1 nHn 1 n 1 n + 1 + 2 1 n + 1 = 2 Hnn 1 n(n + 1) + 1 n + 1 = 2 Hnn n + 1 n(n + 1) = 2 Hnn 1n.

12. [M10] Evaluate H1000 correct to at least 100 decimal places.

By definition

H1000 = k1 1 k1000 = 1 + k2 1 k1000 = 1 + 𝜖

where 𝜖 210001 2100011 1 from exercise 3, and

𝜖 210001 210001 1 1 = 2999 2999 1 1 = 2999 1 + 1 2999 1 1 = 1 2999 1 + 1 1 = 1 2999 1 < 1 2998 = 1 10998 ln 2 ln 10 < 1 10300

so that

H1000 = 1.000

to at least 100 decimal places.

13. [M22] Prove the identity

k=1nxk k = Hn + k=1nn k (x 1)k k .

(Note in particular the special case x = 0, which gives us an identity related to exercise 1.2.6-48.)

Proposition. 1knxk k = Hn + 1knn k (x1)k k .

Proof. Let n be an arbitrary positive integer and x an arbitrary real. We must show that

1knxk k = Hn + 1knn k (x 1)k k .

In the case that n = 1

1k1xk k = x = 1 + 1 1(x 1) = H1 + 1k1 1 k (x 1)k k .

Then, assuming

1knxk k = Hn + 1knn k (x 1)k k

we must show that

1kn+1xk k = Hn+1 + 1kn+1n + 1 k (x 1)k k .

But

1kn+1xk k = 1knxk k + xn+1 n + 1 = 1knxk k + 1 n + 1 + xn+1 n + 1 1 n + 1 = 1knxk k + 1 n + 1 + 1 n + 1 (1 + (x 1))n+1 1 = 1knxk k + 1 n + 1 + 1 n + 1 0kn+1n + 1 k (x 1)k 1 = 1knxk k + 1 n + 1 + 1 n + 1 0kn+1n + 1 k (x 1)k n + 1 0 (x 1)0 = 1knxk k + 1 n + 1 + 1 n + 1 1kn+1n + 1 k (x 1)k = 1knxk k + 1 n + 1 + 1 n + 1 1kn+1n + 1 k n k 1(x 1)k from Eq. 1.2.6-(7) = 1knxk k + 1 n + 1 + 1kn+1n + 1 n + 1 1 k n k 1(x 1)k = 1knxk k + 1 n + 1 + 1kn+1 n k 1 (x 1)k k = 1knxk k + 1 n + 1 + n n + 1 (x 1)n+1 n + 1 + 1kn+1 n k 1 (x 1)k k = Hn + 1knn k (x 1)k k + 1 n + 1 + n n + 1 (x 1)n+1 n + 1 + 1kn+1 n k 1 (x 1)k k = Hn+1 + 1kn+1n k (x 1)k k + 1kn+1 n k 1 (x 1)k k = Hn+1 + 1kn+1 n k + n k 1 (x 1)k k = Hn+1 + 1kn+1n + 1 k (x 1)k k

as we needed to show. □

14. [M22] Show that k=1nHkk = 1 2(Hn2 + H n(2)), and evaluate k=1nHk(k + 1).

We may prove the identity.

Proposition. 1knHkk = 1 2 Hn2 + H n(2).

Proof. Let n be an arbitrary nonnegative integer. We must show that

1knHkk = 1 2 Hn2 + H n(2) .

But

1knHkk 1kn1 kHk = 1kn1 k 1jk1 j = 1kn 1jk1 k 1 j = 1 2 1kn1 k2 + 1kn 1 k2 from Eq. 1.2.3-(13) = 1 2 Hn2 + H n(2)

as we needed to show. □

Thus, we may evaluate the sum as

1knHk(k + 1) = 1kn 1 k + 1Hk = 1kn 1 k + 1 1jk1 j = 1kn 1jk 1 k + 1 1 j = 2kn+1 1jk1 1 k 1 j = 2kn+1 1 k 1 k + 1jk1 j = 2kn+1 1 k 1 k + 2kn+1 1 k 1jk1 j = 2kn+1 1 k2 + 2kn+1 1 k 1jk1 j = 1 + 1kn+1 1 k2 + 2kn+1 1 k 1jk1 j = 1 + Hn+1(2) + 2kn+1 1 k 1jk1 j = 1 Hn+1(2) + 2kn+1 1 k 1jk1 j = 1 Hn+1(2) + 1kn+1 1 k 1jk1 j 1 1 1j11 j = 1 Hn+1(2) + 1kn+1 1 k 1jk1 j 1 = 1 Hn+1(2) + 1kn+1 1 kHk 1 = Hn+1(2) + 1 2 Hn+12 + H n+1(2) = Hn+1(2) + 1 2Hn+12 + 1 2Hn+1(2) = Hn+1(2) + 1 2Hn+12 + 1 2Hn+1(2) = 1 2Hn+12 1 2Hn+1(2) = 1 2 Hn+12 H n+1(2) .

15. [M23] Express k=1nHk2 in terms of n and Hn.

The sum is

1knHk2 = 1knHkHk = 1knHk 1jk1 j = 1kn 1jkHk1 j = 1jn jknHk1 j = 1jn1 j jknHk = 1jn1 j 1knHk 1kj1Hk = 1jn1 j (n + 1)Hn n (j 1 + 1)Hj1 (j 1) from Eq. (8) = 1jn1 j (n + 1)Hn n jHj1 + j 1 = (n + 1)Hn n 1 1jn1 j 1jn1 jjHj1 + 1jn1 jj = (n + 1)Hn n 1Hn 1jnHj1 + 1jn1 = (n + 1)Hn2 nH n Hn 1jnHj Hn + n = (n + 1)Hn2 nH n Hn (n + 1)Hn n Hn + n = (n + 1)Hn2 nH n Hn (n + 1)Hn + n + Hn + n = (n + 1)Hn2 nH n (n + 1)Hn + 2n = (n + 1)Hn2 (2n + 1)H n + 2n.

16. [18] Express the sum 1 + 1 3 + + 1 2n1 in terms of harmonic numbers.

The sum of all n unit fractions with odd denominators through 2n 1 may be expressed as

1kn 1 2k 1 = 1k2n1 k odd 1 k = 1k2n1 1 k 1k2n1 k even 1 k = 1k2n1 1 k 1kn1 1 2k = H2n1 1 2 1kn1 1 k = H2n1 1 2Hn1.

17. [M24] (E. Waring, 1782.) Let p be an odd prime. Show that the numerator of Hp1 is divisible by p.

Proposition. If p is an odd prime, the numerator of Hp1 is divisible by p.

Proof. Let p be an arbitrary odd prime. We must show that the numerator of Hp1 is divisible by p. That is, that

(p 1)!Hp1 = 1kp1(p 1)! k 0(modp).

From exercise 1.2.4-19, the law of inverses, we may find a k such that

kk 1(modp)

since k p. Note that 1 k p 1 and that each k is unique such that {k|1 k p 1} = {k|kk 1(modp)}. Also note that since p is an odd prime by hypothesis, (p1) 2 is an integer. Then, from Wilson’s theorem

(p 1)! 1(modp)

we have that

1kp1(p 1)! k 1kp1 1 k 1kp1kk k 1kp1k 1kp1k p(p 1) 2 0(modp)

as we needed to show. □

______________________________________________________________________________________________________________________________

[Hardy and Wright, An Introduction to the Theory of Numbers, Section 7.8]

18. [M33] (J. Selfridge.) What is the highest power of 2 that divides the numerator of 1 + 1 3 + + 1 2n1?

We want to find the highest power of 2 that divides the numerator of

1kn 1 2k 1,

assuming n positive.

Let m be the integer such that n = 2rm for some integer r. We know that m exists and is odd, as it is the product of the odd primes from the prime factorization of n.

We then have

1kn 1 2k 1 = 1k2rm 1 2k 1 = 0jm1 1k2r 1 j2r+1 + 2k 1,

which we may prove by induction on m.

If m = 1,

1k2r 1 2k 1 = 1k2r 1 (0)2r+1 + 2k 1 = 0j0 1k2r 1 j2r+1 + 2k 1.

Then, assuming

1k2rm 1 2k 1 = 0jm1 1k2r 1 j2r+1 + 2k 1,

we must show that

1k2r(m+1) 1 2k 1 = 0jm 1k2r 1 j2r+1 + 2k 1.

But

1k2r(m+1) 1 2k 1 = 1k2rm 1 2k 1 + 2rm+1k2r(m+1) 1 2k 1 = 0jm1 1k2r 1 j2r+1 + 2k 1 + 2rm+1k2r(m+1) 1 2k 1 = 0jm1 1k2r 1 j2r+1 + 2k 1 + 1k2rm2r 1 2k 1 = 0jm1 1k2r 1 j2r+1 + 2k 1 + 1k2r 1 2(k + 2rm) 1 = 0jm1 1k2r 1 j2r+1 + 2k 1 + 1k2r 1 m2r+1m + 2k 1 = 1k2rm 1 2k 1

and hence the identity.

Let

Pj = 1k2rj2r+1 + 2k 1

be the common denominator such that

1k2r 1 j2r+1 + 2k 1 = 1k2r Pj Pj(j2r+1 + 2k 1) = 1k2rPj(j2r+1 + 2k 1) Pj .

That is, such that the numerator of 1kn 1 2k1 is

0jm1 1k2r Pj j2r+1 + 2k 1,

m sets of 2r terms, each of the form Pj over a distinct odd residue of 2r+1. Each ratio itself is an integer and a distinct odd residue of 2r+1, and the sum of 2r distinct odd residues is 2r2 mj = 22rmj for some integer mj by the odd number theorem, mj odd. That is, the numerator of 1kn 1 2k1 is

0jm122rm j = 22r 0jm1mj.

Since m is odd, we know the sum of m odd terms mj is itself an odd number. Let this be M, so that the numerator of 1kn 1 2k1 is

22rM

for some odd integer M. That is, 22r is the highest power of 2 that divides the numerator of

1kn 1 2k 1

where m is the odd integer such that n = 2rm for some integer r.

________________________________________________________________________________

[AMM 67 (1960), 924–925]

19. [M30] List all nonnegative integers n for which Hn is an integer. [Hint: If Hn has odd numerator and even denominator, it cannot be an integer.]

The nonnegative integers n for which Hn is an integer are n = 0 and n = 1, since H0 = 0 and H1 = 1. To see why these are the only n, consider the following. Let k = lg n with n 2, so that 2k n < 2k+1 and k 1, and let

P = 1in2km

be the common denominator for each term of Hn, m odd but P even. We know that m exists and is odd, as it is the product of the odd primes from a prime factorization of the common denominator. Then

Hn = 1jn1 j = 1jn P Pj = 1jnPj P = 1jnPjP.

If Hn is an integer, then so is Hn 1. But

Hn 1 = 2jnPjP = M P

for M = 2jnPj. Each term in M is even except for the term with j = 2k, P2k = m, which means M is odd. But the divisor P is even. This means their ratio cannot possibly be an integer, and hence the claim.

20. [HM22] There is an analytic way to approach summation problems such as the one leading to Theorem A in this section: If f(x) = k0akxk, and this series converges for x = x0, prove that

k0akx0kH k =01f(x0) f(x0y) 1 y dy.

Proposition. If f(x) = k0akxk and f(x) converges for x = x0 then k0akx0kHk =01f(x0)f(x0y) 1y dy.

Proof. Let f(x) = k0akxk be a series with arbitrary coefficients ak such that f(x) converges for x = x0. We must show that

k0akx0kH k =01f(x0) f(x0y) 1 y dy.

But

k0akx0kH k = k0akx0k 1jk1 j = k0akx0k 1jk01yj1dy = k0akx0k01 1jkyj1dy = k0akx0k01 0jk1yjdy = k0akx0k01y0 yk1+1 1 y dy = k0akx0k011 yk 1 y dy =01 1 1 y k0(akx0k a kx0kyk)dy =01 1 1 y k0akx0k k0ak(x0y)k dy =01 1 1 y f(x0) f(x0y)dy =01f(x0) f(x0y) 1 y dy

as we needed to show. □

______________________________________________________________________________________________________________________________

[AMM 69 (1962), 239; H. W. Gould, Mathematics Magazine 34 (1961), 317–321]

21. [M24] Evaluate k=1nHk(n + 1 k).

The difference between the sum for n and n + 1 is given as

1kn+1 Hk n + 2 k 1kn Hk n + 1 k = Hn+1 n + 2 (n + 1) + 1kn Hk n + 2 k 1kn Hk n + 1 k = Hn+1 + 1kn Hk n + 2 k 1kn Hk n + 1 k = Hn+1 + 1kn Hk n + 2 k Hk n + 1 k = Hn+1 + 1kn Hk n + 2 k 1kn Hk n + 1 k = Hn+1 + 0kn1 Hk+1 n + 1 k 1kn Hk n + 1 k = Hn+1 + H1 n + 1 Hn+1 + 1kn Hk+1 n + 1 k 1kn Hk n + 1 k = 1 n + 1 + 1kn Hk+1 n + 1 k 1kn Hk n + 1 k = 1 n + 1 + 1knHk + 1 k+1 n + 1 k 1kn Hk n + 1 k = 1 n + 1 + 1kn Hk n + 1 k + 1kn 1 (k + 1)(n + 1 k) 1kn Hk n + 1 k = 1 n + 1 + 1kn 1 (k + 1)(n + 1 k) = 1 n + 1 + 1kn 1 (n + 2)(k + 1) + 1 (n + 2)(n + 1 k) = 1 n + 1 + 1 n + 2 1kn 1 k + 1 + 1 n + 1 k = 1 n + 1 + 1 n + 2 1kn 1 k + 1 + 1 n + 2 1kn 1 n + 1 k = 1 n + 1 + 1 n + 2 2kn+1 1 k + 1 n + 2 1kn1 k = 1 n + 1 + 1 n + 2 1 1 + 1 n + 1 + 1kn1 k + Hn n + 2 = 1 n + 1 + 1 n + 2 1 1 + 1 n + 1 + Hn + Hn n + 2 = 1 n + 1 + 1 (n + 1)(n + 2) + Hn 1 n + 2 + Hn n + 2 = 1 n + 1 + Hn n + 2 1 n + 2 + Hn+1 n + 2 = 1 (n + 1)(n + 2) + Hn n + 2 + Hn+1 n + 2 = Hn+1 n + 2 + Hn+1 n + 2 = 2Hn+1 n + 2 .

Then, in the case that n = 1

1k1 Hk 1 + 1 k = Hk 1 + 1 k = H1 = 1 = 9 4 5 4 = 1 + 1 22 1 + 1 22 = H22 H 2(2)

and assuming

1kn Hk n + 1 k = Hn+12 H n+1(2)

it may be shown that

1kn+1 Hk n + 2 k = Hn+22 H n+2(2)

as

1kn+1 Hk n + 2 k = 1kn Hk n + 1 k + 2Hn+1 n + 2 = Hn+12 H n+1(2) + 2Hn+1 n + 2 = Hn+12 + 2Hn+1 n + 2 + 1 (n + 2)2 Hn+1(2) 1 (n + 2)2 = Hn+1 + 1 n + 22 H n+1(2) + 1 (n + 2)2 = Hn+22 H n+2(2).

That is

1kn Hk n + 1 k = Hn+12 H n+1(2).

22. [M28] Evaluate k=0nHkHnk.

From summation by parts and exercise 21,

0knHkHnk = 1knHkHnk = 1jnHj Hnn 1kn 1jkHj Hn(k+1) Hnk = (n + 1)Hn nH0 + 1kn1 (k + 1)Hk k 1 n k = 1kn1(k + 1)Hk k n k = 1kn1(n k + 1)Hnk n + k k = 1kn1nHnk kHnk + Hnk n + k k = 1kn1nHnk k 1kn1kHnk k + 1kn1Hnk k 1kn1n k + 1kn1k k = n 1kn1Hnk k 1kn1Hnk + 1kn1Hnk k n 1kn1 1 k + 1kn11 = n 1kn1Hnk k 1kn1Hnk + 1kn1Hnk k nHn1 + n 1 = n 1kn1 Hk n k 1kn1Hk + 1kn1 Hk n k n(Hn 1) = (n + 1) 1kn1 Hk n k ((n 1) + 1)Hn1 (n 1) n(Hn 1) = (n + 1) 1kn1 Hk n k nHn1 + n 1 n(Hn 1) = (n + 1) 1kn1 Hk n k n(Hn 1) n(Hn 1) = (n + 1) 1kn1 Hk n k 2n(Hn 1) = (n + 1) Hn2 H n(2) 2n(H n 1).

23. [HM20] By considering the function Γ(x)Γ(x), show how we can get a natural generalization of Hn to noninteger values of n. You may use the fact that Γ(1) = γ, anticipating the next exercise.

We can get a natural generalization of Hn to noninteger values of n by considering the function Γ(x)Γ(x), using the fact that Γ(1) = γ.

By definition,

Γ(x + 1) = xΓ(x),

and so

Γ(x + 1) = xΓ(x) = xΓ(x) + xΓ(x) = Γ(x) + xΓ(x)

if and only if

Γ(x + 1) Γ(x + 1) = Γ(x) + xΓ(x) Γ(x + 1) = Γ(x) + xΓ(x) xΓ(x) = Γ(x) xΓ(x) + xΓ(x) xΓ(x) = 1 x + Γ(x) Γ(x) ,

giving us a natural generalization of Hn to noninteger values of n as

Hx = Γ(x + 1) Γ(x + 1) + γ.

Note that in the case that x = 0

H0 = Γ(1) Γ(1) + γ = γ 1 + γ = 0,

and assuming

Hx = Γ(x + 1) Γ(x + 1) + γ

we have that

Hx+1 = Hx + 1 x + 1 = Γ(x + 1) Γ(x + 1) + γ + 1 x + 1 = Γ(x + 1) Γ(x + 1) + 1 x + 1γ = Γ(x + 2) Γ(x + 2) + γ,

proving the identity holds for all nonnegative integers x.

24. [HM21] Show that

xeγx k1 1 + x kexk = 1 Γ(x).

(Consider the partial products of this infinite product.)

Proposition. xeγx k1 1 + x k exk = 1 Γ(x).

Proof. Let x be an arbitrary real. We must show that

xeγx k1 1 + x kexk = 1 Γ(x).

But since

γ = lim nHn ln n

we have that

xeγx k1 1 + x kexk = xeγx lim n 1kn 1 + x kexk = lim nxeγx 1kn 1 + x kexk = lim nxe(Hnln n)x 1kn 1 + x kexk = lim nx exHn ex ln n 1kn 1 + x kexk = lim nxexHn nx 1kn 1 + x kexk = lim n x nxexHn 1kn 1 + x kexk = lim n x nxexHn 1kn 1 + x k 1knexk = lim n x nxexHn 1kn 1 + x kex 1kn1k = lim n x nxexHn 1kn 1 + x kexHn = lim n x nx 1kn 1 + x k = lim n x nx 1knx + k k = lim n x nx 1kn(x + k) 1knk = lim n x nx 1kn(x + k) n! = lim nx 1kn(x + k) nxn! = 1 Γ(x)

as we needed to show. □