Solution to the monotonicity problem of an interesting class of sequences of real numbers
- Open Access
- 04-12-2025
- Research
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Abstract
1 Introduction
Let \({\mathbb{N}}\) be the set of positive natural numbers, \({\mathbb{Z}}\) be the set of whole numbers, \({\mathbb{N}}_{m}=\{l\in {\mathbb{Z}}: l\ge m\}\) for some \(m\in {\mathbb{Z}}\), \({\mathbb{R}}\) be the set of reals, and let \(C_{j}^{n}\), \(0\le j\le n\), be the binomial coefficients ([2, 15, 21]).
Monotone sequences have been investigated for a long time. One can find many examples of such sequences in [1‐10, 13‐22, 24‐37, 39‐41] and in the related references therein. They are frequently connected with some comparison results [5‐7, 26, 31‐35] and can be employed in showing the existence of some specific solutions to recursive relations, including monotone ones, or appear in pairs which are investigated simultaneously [2, 5, 7‐10, 14‐21, 23, 25‐29, 31‐41].
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In [39] was posed a problem, which can be solved by investigating the monotonicity of the sequences therein ([36, 40]). Motivated by the problem, in [36] we investigated monotonicity character of their convex combinations, and gave two solutions to the problem, one of which employed the Hadamard-Hermite inequalities ([11, 12, 20]) and has some interesting geometrical interpretations. The investigation in [36] motivated us to conduct related investigations of some other known sequences of real numbers which appear in pairs, or are somehow connected with the sequences investigated therein.
Motivated by Problems 168 and 172 in [26, Chap. 1], recently in [37] we have studied the monotonicity character of the following two-parameter class of sequences where \(p\ge 0\) and \(\alpha >-1\), whose special cases frequently occur in the literature ([2, 10, 14‐19, 21, 26‐28, 41]). It is clear that for \(p, \alpha \in {\mathbb{R}}\).
$$\begin{aligned} a_{n}^{(p)}(\alpha )=\bigg(1+\frac{\alpha }{n}\bigg)^{n+p},\quad n \in {\mathbb{N}}, \end{aligned}$$
(1)
$$\begin{aligned} \lim _{n\to \infty}a_{n}^{(p)}(\alpha )=e^{\alpha }, \end{aligned}$$
(2)
The following theorem was proved in [37].
Theorem 1
Let \(k\in {\mathbb{N}}\) and \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\) be the sequence defined in (1), where \(p\ge 0\) and \(\alpha >0\). Then the following statements hold.
1.
The sequence satisfies the condition for every \(n\ge k\), if and only if
$$\begin{aligned} a_{n}^{(p)}(\alpha )\le e^{\alpha }, \end{aligned}$$
(3)
$$ p\le \frac{\alpha }{\ln (1+\frac{\alpha }{k})}-k. $$
2.
The sequence satisfies the condition for every \(n\in {\mathbb{N}}\), if and only if \(\alpha \le 2p\).
$$ e^{\alpha }\le a_{n}^{(p)}(\alpha ), $$
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Remark 1
Note that if or equivalently if \((1+p)\ln (1+\alpha )\le \alpha \), Theorem 1 says that \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\) satisfies (3) for every \(n\in {\mathbb{N}}\). Hence, since the relation in (2) holds, of some interest is to know the monotonicity character of the sequence under the condition (4).
$$\begin{aligned} p\le \frac{\alpha }{\ln (1+\alpha )}-1, \end{aligned}$$
(4)
In [37], was also proved that if \(\alpha >0\) and condition (4) holds, then it must be \(2p<\alpha \). On the other hand, the following result, which is an extension of the result in [26, Problem 168, Chap. 1], was also proved in [37].
Theorem 2
Let \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\) be the sequence defined in (1), where \(p\ge 0\) and \(\alpha \in {\mathbb{R}}\setminus \{0\}\). Then, the sequence is strictly decreasing if and only if \(0<\alpha \le 2p\).
Hence, of an interest is the case \(0\le 2p<\alpha \). Regarding the monotonicity character of the sequence (1) in this case, the following result was proved in [37].
Theorem 3
Assume that \(\alpha >0\) and Then, the sequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\) strictly increasingly converges to \(e^{\alpha }\).
$$ 0\le p\le \min \bigg\{ \frac{3\alpha }{\alpha +6}, \frac{2\ln (1+\frac{\alpha }{2})-\ln (1+\alpha )}{\ln (1+\alpha )-\ln (1+\frac{\alpha }{2})} \bigg\} . $$
Theorem 3 gives some sufficient conditions for guarantying the strict monotonicity of the sequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\) in the case \(0\le 2p<\alpha \). Here we give a complete description of the monotonicity character of the sequence in this case, solving the most difficult case, as well as the problem on the monotonicity character of the sequence completely. To do this, among other things, we use several interesting new analytic inequalities.
2 Main results
In this section we prove our main results, which are incorporated in four auxiliary results (i.e., Lemmas 3-6 below), and in a theorem (Theorem 4 below). Before this, we quote two auxiliary results which were proved in [37].
Lemma 1
Let Then, the function f is positive, strictly increasing and concave on the interval \((0,+\infty )\), and satisfies the condition for \(t>0\).
$$ f(t)=\frac{t}{\ln (1+t)}-1. $$
$$ f(t)< \frac{t}{2}, $$
Lemma 2
The following inequality holds for \(t>0\).
$$ \frac{t}{\ln (1+t)}-1> \frac{2\ln (1+\frac{t}{2})-\ln (1+t)}{\ln (1+t)-\ln (1+\frac{t}{2})}, $$
Lemma 3
Let \(k\in {\mathbb{N}}_{2}\) and Then for every \(t>0\) and each \(k\in {\mathbb{N}}_{2}\), \(f_{k}\) is a strictly increasing function on the interval \((0,+\infty )\), and for \(t>0\).
$$ f_{k-1}(t)= \frac{k\ln (1+\frac{t}{k})-(k-1)\ln (1+\frac{t}{k-1})}{\ln (1+\frac{t}{k-1})-\ln (1+\frac{t}{k})}. $$
$$\begin{aligned} 0< f_{k-1}(t)< f_{k}(t), \end{aligned}$$
(5)
$$\begin{aligned} \lim _{k\to +\infty}f_{k}(t)=\frac{t}{2}, \end{aligned}$$
(6)
Proof
It is clear that for every \(t>0\) and each \(k\in {\mathbb{N}}_{2}\).
$$\begin{aligned} \ln \Big(1+\frac{t}{k-1}\Big)>\ln \Big(1+\frac{t}{k}\Big), \end{aligned}$$
(7)
On the other hand, we have and From this, and by some calculation and application of the Bernoulli inequality for \(x>-1\) and \(\alpha \ge 1\) ([2, p.70]), we have for \(2\le j\le k-1\).
$$\begin{aligned} \Big(1+\frac{t}{k-1}\Big)^{k-1}=\sum _{j=0}^{k-1} \frac{C^{k-1}_{j}}{(k-1)^{j}}t^{j}=\sum _{j=0}^{k-1} \frac{(k-1)\cdots (k-j)}{j!(k-1)^{j}}t^{j}, \end{aligned}$$
(8)
$$\begin{aligned} \Big(1+\frac{t}{k}\Big)^{k}=\sum _{j=0}^{k}\frac{C^{k}_{j}}{k^{j}}t^{j}= \sum _{j=0}^{k}\frac{k(k-1)\cdots (k-j+1)}{j!k^{j}}t^{j}. \end{aligned}$$
(9)
$$ (1+x)^{\alpha }\ge 1+\alpha x, $$
$$\begin{aligned} \frac{C^{k}_{j}}{k^{j}}-\frac{C^{k-1}_{j}}{(k-1)^{j}}=& \frac{(k-1)\cdots (k-j+1)}{j!}\Big(\frac{1}{k^{j-1}}- \frac{k-j}{(k-1)^{j}}\Big) \\ =&\frac{(k-1)\cdots (k-j+1)}{j!(k-1)^{j-1}}\Big(\Big(1-\frac{1}{k} \Big)^{j-1}-\frac{k-j}{k-1}\Big) \\ \ge &\frac{(k-1)\cdots (k-j+1)}{j!(k-1)^{j-1}}\Big(1-\frac{j-1}{k}- \frac{k-j}{k-1}\Big) \\ =&\frac{(k-1)\cdots (k-j+1)(j-1)}{j!k(k-1)^{j}}>0, \end{aligned}$$
(10)
For \(j=0,1\), we have
$$\begin{aligned} \frac{C^{k}_{j}}{k^{j}}=\frac{C^{k-1}_{j}}{(k-1)^{j}}. \end{aligned}$$
(11)
From (8)-(11), we easily get for every \(t>0\) and each \(k\in {\mathbb{N}}_{2}\).
$$\begin{aligned} k\ln \Big(1+\frac{t}{k}\Big)>(k-1)\ln \Big(1+\frac{t}{k-1}\Big), \end{aligned}$$
(12)
Let where \(k\in {\mathbb{N}}_{3}\).
$$\begin{aligned} g_{k}(t):=& \frac{k\ln (1+\frac{t}{k})-(k-1)\ln (1+\frac{t}{k-1})}{\ln (1+\frac{t}{k-1})-\ln (1+\frac{t}{k})} \\ &- \frac{(k-1)\ln (1+\frac{t}{k-1})-(k-2)\ln (1+\frac{t}{k-2})}{\ln (1+\frac{t}{k-2})-\ln (1+\frac{t}{k-1})}, \end{aligned}$$
Then
$$\begin{aligned} g_{k}(t)=& \frac{(k\ln (1+\frac{t}{k})-(k-1)\ln (1+\frac{t}{k-1}))(\ln (1+\frac{t}{k-2})-\ln (1+\frac{t}{k-1}))}{(\ln (1+\frac{t}{k-1})-\ln (1+\frac{t}{k}))(\ln (1+\frac{t}{k-2})-\ln (1+\frac{t}{k-1}))} \\ &- \frac{(\ln (1+\frac{t}{k-1})-\ln (1+\frac{t}{k}))((k-1)\ln (1+\frac{t}{k-1})-(k-2)\ln (1+\frac{t}{k-2}))}{(\ln (1+\frac{t}{k-1})-\ln (1+\frac{t}{k}))(\ln (1+\frac{t}{k-2})-\ln (1+\frac{t}{k-1}))}. \end{aligned}$$
(13)
Let for \(k\in {\mathbb{N}}_{3}\).
$$\begin{aligned} &\widehat{g}_{k}(t):= \\ &\Big(k\ln \Big(1+\frac{t}{k}\Big)-(k-1)\ln \Big(1+\frac{t}{k-1}\Big) \Big)\Big(\ln \Big(1+\frac{t}{k-2}\Big)-\ln \Big(1+\frac{t}{k-1}\Big) \Big)- \\ &\Big(\ln \Big(1+\frac{t}{k-1}\Big)-\ln \Big(1+\frac{t}{k}\Big)\Big) \Big((k-1)\ln \Big(1+\frac{t}{k-1}\Big)-(k-2)\ln \Big(1+\frac{t}{k-2} \Big)\Big), \end{aligned}$$
Then, we have and for \(k\in {\mathbb{N}}_{3}\).
$$\begin{aligned} \widehat{g}_{k}(0)=0 \end{aligned}$$
(14)
$$\begin{aligned} \widehat{g}'_{k}(t)= \frac{(k+t)^{2}\ln (1+\frac{t}{k})-2(k-1+t)^{2}\ln (1+\frac{t}{k-1})+(k-2+t)^{2}\ln (1+\frac{t}{k-2})}{(k-2+t)(k-1+t)(k+t)}, \end{aligned}$$
(15)
Let Then, we have and and consequently and from which it follows that and for \(t>0\).
$$\begin{aligned} h_{k}(t):=&(k+t)^{2}\ln \Big(1+\frac{t}{k}\Big)-2(k-1+t)^{2}\ln \Big(1+ \frac{t}{k-1}\Big)+(k-2+t)^{2}\ln \Big(1+\frac{t}{k-2}\Big). \end{aligned}$$
$$\begin{aligned} h_{k}(0)=0 \end{aligned}$$
(16)
$$\begin{aligned} h_{k}'(t)=&2(k+t)\ln \Big(1+\frac{t}{k}\Big)-4(k-1+t)\ln \Big(1+ \frac{t}{k-1}\Big) +2(k-2+t)\ln \Big(1+\frac{t}{k-2}\Big), \end{aligned}$$
$$\begin{aligned} h_{k}'(0)=0 \end{aligned}$$
(17)
$$ h_{k}''(t)=2\ln \Big(1+\frac{t}{k}\Big)-4\ln \Big(1+\frac{t}{k-1} \Big)+2\ln \Big(1+\frac{t}{k-2}\Big), $$
$$\begin{aligned} h_{k}''(0)=0 \end{aligned}$$
(18)
$$\begin{aligned} h_{k}'''(t)=\frac{4}{(k-2+t)(k-1+t)(k+t)}>0, \end{aligned}$$
(19)
From (18) and (19) we have From (17) and (20), it follows that From (16) and (21), it follows that From (15) and (22), it follows that From (7), (13), (14) and (23), it follows that that is, we have for \(t>0\) and \(k\in {\mathbb{N}}_{3}\), which is the second inequality in (5).
$$\begin{aligned} h_{k}''(t)>0\quad \text{ for }\quad t>0. \end{aligned}$$
(20)
$$\begin{aligned} h_{k}'(t)>0\quad \text{ for }\quad t>0. \end{aligned}$$
(21)
$$\begin{aligned} h_{k}(t)>0\quad \text{ for }\quad t>0. \end{aligned}$$
(22)
$$\begin{aligned} \widehat{g}_{k}'(t)>0\quad \text{ for }\quad t>0. \end{aligned}$$
(23)
$$ g_{k}(t)>0\quad \text{ for }\quad t>0, $$
$$ \frac{k\ln (1+\frac{t}{k})-(k-1)\ln (1+\frac{t}{k-1})}{\ln (1+\frac{t}{k-1})-\ln (1+\frac{t}{k})}> \frac{(k-1)\ln (1+\frac{t}{k-1})-(k-2)\ln (1+\frac{t}{k-2})}{\ln (1+\frac{t}{k-2})-\ln (1+\frac{t}{k-1})} $$
Further, we have Let Then we have and for \(t>0\).
$$\begin{aligned} f_{k-1}'(t)= \frac{(k-1+t)\ln (1+\frac{t}{k-1})-(k+t)\ln (1+\frac{t}{k})}{(k-1+t)(k+t)(\ln (1+\frac{t}{k-1})-\ln (1+\frac{t}{k}))^{2}}. \end{aligned}$$
(24)
$$ \widehat{h}_{k}(t)=(k-1+t)\ln \Big(1+\frac{t}{k-1}\Big)-(k+t)\ln \Big(1+ \frac{t}{k}\Big). $$
$$\begin{aligned} \widehat{h}_{k}(0)=0 \end{aligned}$$
(25)
$$\begin{aligned} \widehat{h}_{k}'(t)=\ln \Big(1+\frac{t}{k-1}\Big)-\ln \Big(1+ \frac{t}{k}\Big)>0, \end{aligned}$$
(26)
From (25) and (26) it follows that From (24) and (27), it follows that \(f_{k-1}'(t)>0\) for \(t>0\) which implies strict monotonicity of the function \(f_{k-1}\) on the interval \((0+\infty )\), for every \(k\in {\mathbb{N}}_{2}\).
$$\begin{aligned} \widehat{h}_{k}(t)>0\quad \text{ for }\quad t>0. \end{aligned}$$
(27)
We have proving (6) and finishing the proof of the lemma. □
$$\begin{aligned} &\lim _{k\to +\infty}f_{k-1}(t)=\lim _{k\to +\infty} \frac{k\ln (1+\frac{t}{k})-(k-1)\ln (1+\frac{t}{k-1})}{\ln (1+\frac{t}{k-1})-\ln (1+\frac{t}{k})} \\ =&\lim _{k\to +\infty} \frac{k(\frac{t}{k}-\frac{t^{2}}{2k^{2}}+\frac{t^{3}}{3k^{3}}+O(\frac{1}{k^{4}}))-(k-1)(\frac{t}{k-1}-\frac{t^{2}}{2(k-1)^{2}}+\frac{t^{3}}{3(k-1)^{3}}+O(\frac{1}{k^{4}}))}{\frac{t}{k-1}-\frac{t^{2}}{2(k-1)^{2}}+O(\frac{1}{k^{3}})-(\frac{t}{k}-\frac{t^{2}}{2k^{2}}+O(\frac{1}{k^{3}}))} \\ =&\lim _{k\to +\infty} \frac{(t-\frac{t^{2}}{2k}+\frac{t^{3}}{3k^{2}}+O(\frac{1}{k^{3}}))-(t-\frac{t^{2}}{2k}(1-\frac{1}{k})^{-1}+\frac{t^{3}}{3k^{2}}(1-\frac{1}{k})^{-2}+O(\frac{1}{k^{3}}))}{\frac{t}{k}(1-\frac{1}{k})^{-1}-\frac{t^{2}}{2k^{2}}(1-\frac{1}{k})^{-2}+O(\frac{1}{k^{3}})-(\frac{t}{k}-\frac{t^{2}}{2k^{2}}+O(\frac{1}{k^{3}}))} \\ =&\lim _{k\to +\infty} \frac{\frac{t^{2}}{2k}((1-\frac{1}{k})^{-1}-1)-\frac{t^{3}}{3k^{2}}((1-\frac{1}{k})^{-2}-1)+O(\frac{1}{k^{3}})}{\frac{t}{k}((1-\frac{1}{k})^{-1}-1)-\frac{t^{2}}{2k^{2}}((1-\frac{1}{k})^{-2}-1)+O(\frac{1}{k^{3}})} \\ =&\lim _{k\to +\infty} \frac{\frac{t^{2}}{2k^{2}}+O(\frac{1}{k^{3}})}{\frac{t}{k^{2}}+O(\frac{1}{k^{3}})}=\frac{t}{2}, \end{aligned}$$
Lemma 4
Let \(k\in {\mathbb{N}}_{2}\) and Then for every \(t>0\) and each \(k\in {\mathbb{N}}_{2}\), and \(g_{k}\) is a strictly increasing function on the interval \((0,+\infty )\).
$$ g_{k}(t)=\frac{kt}{t+2k}. $$
$$\begin{aligned} 0< g_{k-1}(t)< g_{k}(t), \end{aligned}$$
(28)
Proof
It is obvious that \(g_{k}(t)>0\), for every \(t>0\) and each \(k\in {\mathbb{N}}\). Further, we have for \(t>0\) and \(k\in {\mathbb{N}}_{2}\), that is, (28) holds.
$$ g_{k}(t)-g_{k-1}(t)=\frac{kt}{t+2k}-\frac{(k-1)t}{t+2k-2}= \frac{t^{2}}{(t+2k)(t+2k-2)}>0 $$
Since for \(t>0\) and \(k\in {\mathbb{N}}\), we obtain that \(g_{k}\) is a strictly increasing function on the interval \((0,+\infty )\). □
$$ g_{k}'(t)=\frac{2k^{2}}{(t+2k)^{2}}>0 $$
Lemma 5
Let \(k\in {\mathbb{N}}\) and Then, the following statements hold.
$$\begin{aligned} h_{k}(t)= \frac{2\ln (1+\frac{t}{2})-\ln (1+t)}{\ln (1+t)-\ln (1+\frac{t}{2})}- \frac{kt}{t+2k}. \end{aligned}$$
(29)
1.
For each \(k\in {\mathbb{N}}_{3}\) there is a unique positive zero \(t_{k}\) of the function \(h_{k}\), whereas for \(k=1\) and \(k=2\) the function is positive on the interval \((0,+\infty )\).
2.
The sequence \((t_{k})_{k\in {\mathbb{N}}_{3}}\) is strictly increasing and unbounded.
Proof
(a) We have Let Then, we have and and consequently and
$$ h_{k}(t)= \frac{(t+2k)(2\ln (1+\frac{t}{2})-\ln (1+t))-kt(\ln (1+t)-\ln (1+\frac{t}{2}))}{(\ln (1+t)-\ln (1+\frac{t}{2}))(t+2k)}. $$
$$ p_{k}(t)=(t+2k)\Big(2\ln \Big(1+\frac{t}{2}\Big)-\ln (1+t)\Big)-kt \Big(\ln (1+t)-\ln \Big(1+\frac{t}{2}\Big)\Big). $$
$$\begin{aligned} p_{k}(0)=0 \end{aligned}$$
(30)
$$ p_{k}'(t)=(k+2)\ln \Big(1+\frac{t}{2}\Big)-(k+1)\ln (1+t)+1+ \frac{2k-4}{t+2}-\frac{k-1}{t+1}, $$
$$\begin{aligned} p_{k}'(0)=0 \end{aligned}$$
(31)
$$\begin{aligned} p_{k}''(t)=\frac{t(t^{2}-2(k-3)t-3(k-2))}{((t+1)(t+2))^{2}}. \end{aligned}$$
(32)
From (32) we have if \(k=1\) or \(k=2\).
$$\begin{aligned} p_{k}''(t)>0\quad \text{ for }\quad t>0, \end{aligned}$$
(33)
From (31) and (33), it follows that if \(k=1\) or \(k=2\). From (30) and (34), it follows that \(p_{k}(t)>0\) for \(t>0\), and consequently \(h_{k}(t)>0\) for \(t>0\), in this case.
$$\begin{aligned} p_{k}'(t)>0\quad \text{ for }\quad t>0, \end{aligned}$$
(34)
If \(k\in {\mathbb{N}}_{3}\), then we have that and which implies that \(p_{k}'(t)\) decreases on the interval \((0,k-3+\sqrt{k^{2}-3k+3})\) and increases on the interval \((k-3+\sqrt{k^{2}-3k+3},+\infty )\).
$$ p_{k}''(t)< 0\quad \text{ for }\quad t\in (0,k-3+\sqrt{k^{2}-3k+3}) $$
$$ p_{k}''(t)>0\quad \text{ for }\quad t>k-3+\sqrt{k^{2}-3k+3}, $$
From this and since it follows that there is a unique point \(s_{k}\in (k-3+\sqrt{k^{2}-3k+3},+\infty )\) such that \(p_{k}'(t)<0\) for \(t\in (0,s_{k})\) and \(p_{k}'(t)>0\) for \(t>s_{k}\), which implies that \(p_{k}(t)\) decreases on the interval \((0,s_{k})\) and increases on the interval \((s_{k},+\infty )\). From this and since it follows that there is a unique point \(t_{k}\in (s_{k},+\infty )\) such that \(p_{k}(t)<0\) for \(t\in (0,t_{k})\) and \(p_{k}(t)>0\) for \(t\in (t_{k},+\infty )\).
$$ \lim _{t\to +\infty} p_{k}'(t)=+\infty $$
$$ \lim _{t\to +\infty} p_{k}(t)=+\infty $$
(b) Now note that by Lemma 4, we have for \(t>0\).
$$\begin{aligned} h_{k}(t)-h_{k-1}(t)=g_{k-1}(t)-g_{k}(t)< 0, \end{aligned}$$
(35)
Further, we have \(h_{k-1}(t_{k-1})=0\), from which together with (35) it follows that for \(t\in (0,t_{k-1}]\). Since \(h_{k}\) has a unique positive zero \(t_{k}\), from this we have that it must be \(t_{k}>t_{k-1}\).
$$ h_{k}(t)< h_{k-1}(t)\le 0, $$
Now assume that the sequence \((t_{k})_{k\in {\mathbb{N}}_{3}}\) is bounded. Then, there is a positive finite Since by letting \(k\to +\infty \) in (29) we get since the function \(f_{1}\) is continuous and as \(k\to +\infty \).
$$ \lim _{k\to +\infty}t_{k}=t^{*}. $$
$$ f_{1}(t_{k})=g_{k}(t_{k}), $$
$$\begin{aligned} \frac{2\ln (1+\frac{t^{*}}{2})-\ln (1+t^{*})}{\ln (1+t^{*})-\ln (1+\frac{t^{*}}{2})}= \frac{t^{*}}{2}, \end{aligned}$$
(36)
$$\begin{aligned} \left |\frac{kt_{k}}{t_{k}+2k}-\frac{t^{*}}{2}\right |=&\left | \frac{kt_{k}}{t_{k}+2k}-\frac{kt^{*}}{t^{*}+2k}+ \frac{kt^{*}}{t^{*}+2k}-\frac{t^{*}}{2}\right | \\ =&\left |\frac{2k^{2}(t_{k}-t^{*})}{(t_{k}+2k)(t^{*}+2k)}- \frac{(t^{*})^{2}}{2(t^{*}+2k)}\right | \\ \le &\frac{|t_{k}-t^{*}|}{2}+\frac{(t^{*})^{2}}{4k}\to 0, \end{aligned}$$
Lemma 6
The following inequality holds for \(t>0\) and \(k\in {\mathbb{N}}\).
$$\begin{aligned} \frac{t}{\ln (1+\frac{t}{k})}-k> \frac{(k+1)\ln (1+\frac{t}{k+1})-k\ln (1+\frac{t}{k})}{\ln (1+\frac{t}{k})-\ln (1+\frac{t}{k+1})}, \end{aligned}$$
(37)
Proof
Let Then, we have
$$ \widetilde{f}_{k}(t)=\frac{t}{\ln (1+\frac{t}{k})}-k- \frac{(k+1)\ln (1+\frac{t}{k+1})-k\ln (1+\frac{t}{k})}{\ln (1+\frac{t}{k})-\ln (1+\frac{t}{k+1})}. $$
$$\begin{aligned} \widetilde{f}_{k}(t)=& \frac{(t-k\ln (1+\frac{t}{k}))(\ln (1+\frac{t}{k})-\ln (1+\frac{t}{k+1}))}{(\ln (1+\frac{t}{k})-\ln (1+\frac{t}{k+1}))\ln (1+\frac{t}{k})} \\ &- \frac{((k+1)\ln (1+\frac{t}{k+1})-k\ln (1+\frac{t}{k}))\ln (1+\frac{t}{k})}{(\ln (1+\frac{t}{k})-\ln (1+\frac{t}{k+1}))\ln (1+\frac{t}{k})}. \end{aligned}$$
(38)
Let Then, we have and
$$\begin{aligned} g_{k}(t)=&\Big(t-k\ln \Big(1+\frac{t}{k}\Big)\Big)\Big(\ln \Big(1+ \frac{t}{k}\Big)-\ln \Big(1+\frac{t}{k+1}\Big)\Big) \\ &-\Big((k+1)\ln \Big(1+\frac{t}{k+1}\Big)-k\ln \Big(1+\frac{t}{k} \Big)\Big)\ln \Big(1+\frac{t}{k}\Big). \end{aligned}$$
$$\begin{aligned} g_{k}(0)=0 \end{aligned}$$
(39)
$$\begin{aligned} g_{k}'(t)= \frac{(k+t)^{2}\ln (1+\frac{t}{k})+t-(k+1+t)^{2}\ln (1+\frac{t}{k+1})}{(k+t)(k+1+t)}. \end{aligned}$$
(40)
Let Then, we have and and consequently and for \(t>0\).
$$ h_{k}(t)=(k+t)^{2}\ln \Big(1+\frac{t}{k}\Big)-(k+1+t)^{2}\ln \Big(1+ \frac{t}{k+1}\Big)+t. $$
$$\begin{aligned} h_{k}(0)=0 \end{aligned}$$
(41)
$$ h_{k}'(t)=2(k+t)\ln \Big(1+\frac{t}{k}\Big)-2(k+1+t)\ln \Big(1+ \frac{t}{k+1}\Big), $$
$$\begin{aligned} h_{k}'(0)=0 \end{aligned}$$
(42)
$$\begin{aligned} h_{k}''(t)=2\ln \Big(1+\frac{t}{k}\Big)-2\ln \Big(1+\frac{t}{k+1} \Big)>0, \end{aligned}$$
(43)
From (42) and (43) it follows that From (41) and (44), it follows that From (40) and (45), it follows that From (39) and (46), we get From (38) and (47), the inequality in (37) follows. □
$$\begin{aligned} h_{k}'(t)>0\quad \text{ for }\quad t>0, \end{aligned}$$
(44)
$$\begin{aligned} h_{k}(t)>0\quad \text{ for }\quad t>0. \end{aligned}$$
(45)
$$\begin{aligned} g_{k}'(t)>0\quad \text{ for }\quad t>0. \end{aligned}$$
(46)
$$\begin{aligned} g_{k}(t)>0\quad \text{ for }\quad t>0. \end{aligned}$$
(47)
The following result is the main in the paper. It completely describes the monotonicity character of the sequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\).
Theorem 4
Assume that \(\alpha >0\). Then the following statements hold.
1.
If Then, the sequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\) strictly increasingly converges to \(e^{\alpha }\).
$$\begin{aligned} 0\le p< \frac{2\ln (1+\frac{\alpha }{2})-\ln (1+\alpha )}{\ln (1+\alpha )-\ln (1+\frac{\alpha }{2})}. \end{aligned}$$
(48)
2.
If Then, and the subsequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}_{2}}\) strictly increasingly converges to \(e^{\alpha }\).
$$\begin{aligned} p= \frac{2\ln (1+\frac{\alpha }{2})-\ln (1+\alpha )}{\ln (1+\alpha )-\ln (1+\frac{\alpha }{2})}. \end{aligned}$$
(49)
$$\begin{aligned} a_{1}^{(p)}(\alpha )=a_{2}^{(p)}(\alpha ) \end{aligned}$$
(50)
3.
If for some \(k_{0}\in {\mathbb{N}}_{2}\), then the subsequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}_{k_{0}}}\) strictly increasingly converges to \(e^{\alpha }\), whereas the finite subsequence \((a_{n}^{(p)}(\alpha ))_{n=1}^{k_{0}}\) is strictly decreasing.
$$\begin{aligned} f_{k_{0}-1}(\alpha )< p< f_{k_{0}}(\alpha ), \end{aligned}$$
(51)
4.
If for some \(k_{0}\in {\mathbb{N}}_{2}\), then the subsequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}_{k_{0}+1}}\) strictly increasingly converges to \(e^{\alpha }\), whereas the finite subsequence \((a_{n}^{(p)}(\alpha ))_{n=1}^{k_{0}}\) is strictly decreasing.
$$\begin{aligned} p=f_{k_{0}}(\alpha ), \end{aligned}$$
(52)
$$\begin{aligned} a_{k_{0}}^{(p)}(\alpha )=a_{k_{0}+1}^{(p)}(\alpha ) \end{aligned}$$
(53)
Proof
(a) By Lemma 2 and Theorem 1 we see that condition (48) implies that the sequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\) satisfies the inequalities in (3).
By Lemma 3 the sequence is strictly increasing for any \(\alpha >0\). Hence, the condition (48) implies for each \(k\in {\mathbb{N}}_{2}\), which implies that for every \(k\in {\mathbb{N}}_{2}\), which means that the sequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\) is strictly increasing, and as we know it converges to \(e^{\alpha }\).
$$\begin{aligned} \frac{k\ln (1+\frac{\alpha }{k})-(k-1)\ln (1+\frac{\alpha }{k-1})}{\ln (1+\frac{\alpha }{k-1})-\ln (1+\frac{\alpha }{k})}, \quad k\in {\mathbb{N}}_{2}, \end{aligned}$$
(54)
$$\begin{aligned} p< \frac{k\ln (1+\frac{\alpha }{k})-(k-1)\ln (1+\frac{\alpha }{k-1})}{\ln (1+\frac{\alpha }{k-1})-\ln (1+\frac{\alpha }{k})} \end{aligned}$$
(55)
$$ a_{k-1}^{(p)}(\alpha )< a_{k}^{(p)}(\alpha ) $$
(b) By Lemma 2 and Theorem 1 we see that condition (48) implies that the sequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}}\) satisfies the inequalities in (3). The condition (49) is equivalent to (50), whereas the strict monotonicity of the sequence (54) implies that inequality (55) holds for \(k\in {\mathbb{N}}_{3}\), which means that the subsequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}_{2}}\) is strictly increasing.
(c) If (51) holds for some \(k_{0}\in {\mathbb{N}}_{2}\), then by Lemma 6 and Theorem 1 we see that the subsequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}_{k_{0}}}\) satisfies the inequalities in (3). The monotonicity of the sequence (54) implies that (55) holds for \(k\ge k_{0}\), so the subsequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}_{k_{0}}}\) strictly increasing. On the other hand, we have for \(1\le k\le k_{0}\), which means that for \(1\le k\le k_{0}\), that is, the finite subsequence \((a_{n}^{(p)}(\alpha ))_{n=1}^{k_{0}}\) is strictly decreasing.
$$\begin{aligned} p> \frac{k\ln (1+\frac{\alpha }{k})-(k-1)\ln (1+\frac{\alpha }{k-1})}{\ln (1+\frac{\alpha }{k-1})-\ln (1+\frac{\alpha }{k})} \end{aligned}$$
(56)
$$ a_{k-1}^{(p)}(\alpha )>a_{k}^{(p)}(\alpha ) $$
(d) By Lemma 6 and Theorem 1 we see that condition (52) implies that the sequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}_{k_{0}}}\) satisfies the inequalities in (3). The condition (52) is equivalent to (53), whereas the strict monotonicity of the sequence (54) implies that inequality (55) holds for \(k\in {\mathbb{N}}_{k_{0}+2}\), which means that the subsequence \((a_{n}^{(p)}(\alpha ))_{n\in {\mathbb{N}}_{k_{0}+1}}\) is strictly increasing. On the other hand, we have that (56) holds for \(1\le k\le k_{0}\), which means that for \(1\le k\le k_{0}\), that is, the finite subsequence \((a_{n}^{(p)}(\alpha ))_{n=1}^{k_{0}}\) is strictly decreasing. □
$$ a_{k-1}^{(p)}(\alpha )>a_{k}^{(p)}(\alpha ) $$
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