����������� ������ 50 (1998), 5–10

UDK 517.521

���������� ������ ��� research paper

BEST POSSIBLE BOUNDS AND MONOTONICITY OF SEGMENTS OF HARMONIC SERIES (II)

Slavko Simić

Abstract. We give an answer to a hypothesis formulated in one of our earlier papers, concerning boundaries and estimates of some segments of harmonic series.

In our previous article [4] we established some monotonicity criteria for the sequences s(a_{n}, b_{n}) of the type

s(a_{n}, b_{n}) := \frac{1}{a_{n}} + \frac{1}{a_{n} + 1} + \frac{1}{a_{n} + 2} + ⋯ + \frac{1}{b_{n}} ,

where (a_{n}) and (b_{n}) are increasing sequences of positive integers and a_{n} < b_{n}, n ∈ N.

Let A_{n} := (a_{n} − 1/2)^{2} + 1/12, B_{n} := (b_{n} + 1/2)^{2} + 1/12, n ∈ N. Our results from [4] are contained in the following propositions:

Proposition A. If the sequence (A_{n}/B_{n}) is nonincreasing for n ∈ [n_{1}, n_{2}], then s(a_{n}, b_{n}) is strictly decreasing for n ∈ [n_{1}, n_{2}], n_{1}, n_{2} ∈ N.

Proposition B. If the sequence (\frac{B_{n}−1/12}{A_{n}−1/12}), i.e. (\frac{b_{n}+1/2}{a_{n}−1/2}) is nondecreasing, then s(a_{n}, b_{n}) is strictly increasing for the same n ∈ [n_{1}, n_{2}].

Proposition C. If

\frac{B_{n+1}}{A_{n+1}} − \frac{B_{n}}{A_{n}} > \frac{B_{n+1} − B_{n}}{45a_{n}^{2}A_{n}A_{n+1}} , n ∈ [n_{1}, n_{2}),

then the sequence s(a_{n}, b_{n}) is strictly increasing for n ∈ [n_{1}, n_{2}].

There are numerous problems concerning boundaries and estimates of some segments of harmonic series (i.e. s(a_{n}, b_{n})) for given integer sequences (a_{n}), (b_{n}); see for example [1], [2].

AMS Subject Classification: 26 D 15

Keywords and phrases: Harmonic series

Since the best possible bounds for s(a_{n}, b_{n}) are represented by this sequence itself, we can define (as in [2]) the best lower bound of s(a_{n}, b_{n}) by s_{∗} = inf_{n} s(a_{n}, b_{n}), and the best possible upper bound by s^{∗} = sup_{n} s(a_{n}, b_{n}); so these bounds do not depend on n ∈ N.

Obviously, the question of the best possible bounds is in close connection with monotonicity of the sequence s(a_{n}, b_{n}) and Propositions A–C, cited above, give an efficient tool for solving this problem for wide class of segments of harmonic series.

In [3] we considered the question of the best possible bounds for sequences (a_{n}), (b_{n}) being of the form of arithmetic progression, i.e.

s_{n}(p, q, A, B) := s(np + A, nq + B) = \frac{1}{np + A} + \frac{1}{np + A + 1} + ⋯ + \frac{1}{nq + B} ,

where p, q, A, B are fixed integers and q > p > 0, B ⩾ A − 1 ⩾ 0. There we formulated a generalization of results from [2] and [3] in the following

Hypothesis 1. Let m := p(2B + 1) − q(2A − 1). Then the sequence s_{n}(p, q, A, B) is:

(a) strictly increasing if m ⩽ 0; hence the best possible bounds are

s_{∗}(p, q, A, B) = inf_{n} s_{n}(p, q, A, B) = s_{1}(p, q, A, B);
s^{∗}(p, q, A, B) = sup_{n} s_{n}(p, q, A, B) = lim_{n} s_{n}(p, q, A, B) = ln(q/p);

(b) strictly decreasing in the case that m > 0, so

s_{∗}(p, q, A, B) = ln(q/p); s^{∗}(p, q, A, B) = s_{1}(p, q, A, B).

In the same article we proved the validity of the cited hypothesis for the cases A = 1, B = 0, p, q ∈ N, and A = 1, B = 1, p, q ∈ N. We also showed that s_{n}(p, q, A, B) is monotonous for sufficiently large n under conditions on p, q, A, B, m as defined.

But closer examination shows that part (b) of the hypothesis is not valid, i.e. condition m > 0 is necessary but not sufficient for s_{n}(p, q, A, B) to be strictly decreasing for each n, n ∈ N. This can be illustrated by the following example.

Let p = A = 1, q = 2B; then m = p(2B + 1) − q(2A − 1) = 1 > 0, but

s_{n+1}(1, 2B, 1, B)−s_{n}(1, 2B, 1, B) = ∑_{s=n+2}^{B(2n+3)} \frac{1}{s} − ∑_{s=n+1}^{B(2n+1)} \frac{1}{s} = ∑_{s=B(2n+1)+1}^{B(2n+3)} \frac{1}{s} − \frac{1}{n + 1}

> ∫_{B(2n+1)+1}^{B(2n+3)+1} \frac{dt}{t} − \frac{1}{n + 1} = ln (1 + \frac{1}{n + \frac{1}{2} + \frac{1}{2B}} ) − \frac{1}{n + 1} > 0,

for B large enough and fixed n, since the inequality ln (1 + \frac{1}{n + \frac{1}{2}} ) > \frac{1}{n + 1} is satisfied for each n, n ∈ N.

In this paper we will give definite solution of cited Hypothesis using Propositions A–C, i.e. we will show that part (a) is correct and, under conditions of part (b), we will determine n_{0} = n_{0}(m, p, q, A) such that for n ⩽ n_{0} the sequence s_{n}(p, q, A, B) is strictly increasing and for n ⩾ n_{0} otherwise. From this, the best possible bounds follow immediately.

An interesting consequence of this proposition is the establishment of the best possible bounds for a generalized sequence s_{f(n)}(p, q, A, B):

s_{f(n)}(p, q, A, B) := \frac{1}{pf(n) + A} + \frac{1}{pf(n) + A + 1} + ⋯ + \frac{1}{qf(n) + B} ,

where f: N → N ia any increasing integer function.

An answer to cited Hypothesis 1 (which includes all cases considered in [3]) is given in the next

Theorem D. Let A, B, p, q be fixed integers satisfying B ⩾ A − 1 ⩾ 0, q > p ⩾ 1, and define

m := p(2B + 1) − q(2A − 1); c := \frac{q}{6mp} − \frac{2A − 1}{2p}; C_{1} := [c − \frac{1}{3}, c + \frac{1}{3}];

C_{2} := (c + \frac{1}{3}, c + \frac{2}{3}); r ∈ C_{1} ∪ C_{2} is an integer and t := max(1, r).

Considering the sequence s_{n}(p, q, A, B) defined as above, we have:

Proposition D1. If m ⩽ 0, the sequence s_{n}(p, q, A, B) is strictly increasing for each n, n ∈ N; hence the best possible bounds are:

s_{∗}(p, q, A, B) := inf_{n} s_{n}(p, q, A, B) = s_{1}(p, q, A, B) = ∑_{s=p+A}^{q+B} \frac{1}{s};
s^{∗}(p, q, A, B) := sup_{n} s_{n}(p, q, A, B) = lim_{n} s_{n}(p, q, A, B) = ln(q/p).

Proof. Checking validity of conditions of cited Proposition B (for a_{n} = pn+A; b_{n} = qn + B), we obtain that m ⩽ 0 is equivalent to

\frac{nq + B + 1/2}{np + A − 1/2} ⩾ \frac{(n + 1)q + B + 1/2}{(n + 1)p + A − 1/2},

i.e. (\frac{b_{n}+1/2}{a_{n}−1/2}) is nondecreasing. Therefore s_{n}(p, q, A, B) is strictly increasing and conclusions of Proposition D1 follow.

Proposition D2. In the case m > 0, the sequence s_{n}(p, q, A, B) has the maximal term with index n_{0} in the sense that it strictly increases for n ∈ [1, n_{0}] and strictly decreases for n ⩾ n_{0}. Index n_{0} is determined by: (i) n_{0} = t if r ∈ S_{1}; (ii) n_{0} = t or n_{0} = t + 1 if r ∈ S_{2}.

Proof. Let m > 0 and suppose n_{0} ⩾ 2. Considering case (i), for n ∈ [1, n_{0}) we have c − (n + 1/2) ⩾ c − (n_{0} − 1/2) ⩾ 1/6, that is

6mp(c − (n + 1/2)) ⩾ mp > \frac{3m(2A − 1) + 3mp(2n + 1)}{3,75(np + A)^{2} − 1},

wherefrom 22,5(np + A)^{2}mp(c − (n + 1/2)) > q, i.e.

45(np + A)^{2}mpq (c − (n + \frac{1}{2})) (n + \frac{1}{2} + \frac{2B + 1}{2q}) > q((2n + 1)q + 2B + 1),

i.e.

45(np + A)^{2}mpq ((c − (n + \frac{1}{2})) (n + \frac{1}{2} + \frac{2B + 1}{2q} + \frac{p}{6mq} ) + \frac{1}{4} − \frac{1}{36m^{2}} )

> q((2n + 1)q + 2B + 1).

This is (with a_{n} = pn + A, b_{n} = qn + B) exactly

45a_{n}^{2}(B_{n+1}A_{n} − B_{n}A_{n+1}) > B_{n+1} − B_{n};

so we have proved the validity of the conditions of Proposition C. Hence, s_{n}(p, q, A, B) is strictly increasing for n ∈ [1, n_{0}].

Otherwise, for m > 0 and n ⩾ n_{0} we have

B_{n}A_{n+1} − B_{n+1}A_{n} = ((nq + B + \frac{1}{2})^{2} + \frac{1}{12})(((n + 1)p + A − \frac{1}{2})^{2} + \frac{1}{12})−

− (((n + 1)q + B + \frac{1}{2})^{2} + \frac{1}{12})((np + A − \frac{1}{2})^{2} + \frac{1}{12})

= mpq ((n + \frac{1}{2} − c) (n + \frac{1}{2} + \frac{2B + 1}{2q} + \frac{p}{6mq} ) − \frac{1}{4} + \frac{1}{36m^{2}} )

⩾ mpq ((n_{0} + \frac{1}{2} − c) (n_{0} + \frac{1}{2} + \frac{2B + 1}{2q} + \frac{p}{6mq} ) − \frac{1}{4} + \frac{1}{36m^{2}} ) > 0,

since n_{0} + \frac{1}{2} − c \{ \begin{matrix}> 1/2; & c < 1, \\ ⩾ 1/6; & c ⩾ 1. \end{matrix}

Hence, under the conditions of Proposition A, the sequence s_{n}(p, q, A, B) is strictly decreasing for n ⩾ n_{0}. It follows that, in this case:

s^{∗}(p, q, A, B) = s_{n_{0}}(p, q, A, B); s_{∗}(p, q, A, B) = min(s_{1}(p, q, A, B), ln(q/p)).

What is the exact value of this minimum stays ambiguous and depends how large is the ratio p/A (as could be seen using inequalities from [4]).

The proof of the case (ii) goes on similar lines, so we omit it here.

Now we expose a generalized form of Theorem D.

Theorem E. Let f: N → N be any strictly increasing integer function. Define:

S_{f(n)} = S_{f(n)}(p, q, A, B) := \frac{1}{pf(n) + A} + \frac{1}{pf(n) + A + 1} + ⋯ + \frac{1}{qf(n) + B} ,

with p, q, A, B, m, c, r, t, n_{0} same as in Theorem D. Let

S_{∗} = S_{∗}(p, q, A, B) := inf_{n} S_{f(n)}, S^{∗} = S^{∗}(p, q, A, B) := sup_{n} S_{f(n)}.

Proposition E1. If m < 0 then the sequence S_{f(n)} is strictly increasing for each n, n ∈ N; hence, S_{∗} = S_{f(1)}; S^{∗} = S_{f(∞)} = ln(q/p).

Proposition E2. If m > 0, (S_{f(n)}) is strictly increasing for n ∈ [1, k_{0}], where k_{0} := max(1, max_{k}(f(k) ⩽ n_{0})). So, in this case S_{∗} = S_{f(1)}; S^{∗} = S_{f(k_{0})}.

Proposition E3. If m > 0, (S_{f(n)}) is strictly decreasing for n ⩾ k_{1}, where k_{1} := min_{k}(f(k) ⩾ n_{0}); hence: S_{∗} = S_{f(∞)} = ln(q/p); S^{∗} = S_{f(k_{1})}.

Proof of Theorem E is obvious and immediately follows from Theorem D. For example, validity of Proposition E1 could be proved like this:

Let f(i) = k_{i}, i ∈ N. Since f(·) is strictly increasing on N, so is the sequence (k_{i}). According to Proposition D1, for m ⩽ 0, s_{n}(p, q, A, B) is strictly increasing for each n, n ∈ N. We conclude that s_{k_{1}}(·) < s_{k_{2}}(·) < ⋯ < s_{k_{n}}(·) < ⋯, i.e. S_{f(1)} < S_{f(2)} < ⋯ < S_{f(n)} < · · · .

Propositions E2,3 could be proved in a similar manner.

For an illustration, we take sequences (a_{n}), (b_{n}) in the form of geometrical progression, i.e. a_{n} = pa^{n}, b_{n} = qa^{n}; p < q, a > 1; p, q, a ∈ N.

Proposition F. The sequence

s(pa^{n}, qa^{n}) := \frac{1}{pa^{n}} + \frac{1}{pa^{n} + 1} + \frac{1}{pa^{n} + 2} + ⋯ + \frac{1}{qa^{n}}

is strictly decreasing for each n, n ∈ N. Hence, s_{∗} = lim_{n} s(·) = ln(q/p); s^{∗} = s(pa, qa).

Proof. Put in Theorem E: A = p, B = q, f(n) = a^{n} − 1; then: m := p(2B + 1) − q(2A − 1) = p + q > 0; c < 0; n_{0} = 1; so, from Proposition E3 it follows that the considered sequence is strictly decreasing for each n, n ∈ N, and conclusion follows.

For the sake of completeness we note a possibility to prove (using Proposition A), that the sequence (s(pa^{n}, qb^{n})); a < b, pa ⩽ qb; a, b, p, q ∈ N; is strictly increasing for each n, n ∈ N. Hence, in this case s_{∗} = s(pa, qb); s^{∗} = +∞.

REFERENCES

[1] Problem 108, Mat. vesnik 4 (19) (1967).

[2] Adamović, D. and Tasković, M., Monotony and the best possible bounds of some sequences of sums, Publ. Elek. fak. Beograd, 247–275 (1969).

[3] Simić, S., Proof of a hypothesis on some segments of harmonic series and a new hypothesis, Mat. Vesnik 5 (16)(31) (1979).

[4] Simić, S., Monotonicity of segments of harmonic series (I), Mat. vesnik 38 (1986).

(received 22.10.1996.)

Baba Višnjina 34, 11000 Beograd, Yugoslavia