We begin with the following auxiliary result of independent interest.
We will also need the following auxiliary result, which gives bounds for the error of approximation of an integral by the trapezoidal rule in terms of the first derivative of a function. Before stating the result, we note that such a result is known, but only in the asymptotic case (
asymptotic error estimate): denoting by
\(E_{N}(f)\) the exact error and
\(\tilde{E}_{N}(f)=\frac{1}{12 N^{2}}(f'(1)-f'(0))\), we have
\(\lim_{N\rightarrow \infty } E_{N}(f)/\tilde{E}_{N}(f)=1\) (see [
1, Sect. 5.1]), whereas our result below shows that
\(0\le E_{N}(f)\le \frac{1}{4 N^{2}}(f'(1)-f'(0))\) for all
\(N\ge 1\).
The following technical result is essential for the proof of our main results in the following section.
Proof
Under the hypotheses on n and k, it is easy to verify that \(\frac{k-1}{n-1}<\frac{k}{n}<\frac{n-1}{2n-k-2}\leq 1\) (the last inequality is strict if \(k< n-1\)).
If \(k=0\), then since \(\frac{1}{1-x-\frac{i}{n-1}x}>\frac{1}{1- \frac{i}{n-1}x}\) for \(i\in \{ 0,\ldots ,n-1 \} \) and \(x\in (0,\frac{1}{2} )\), we have \(\varphi _{n,0} ( x ) <0\) for \(x\in (0,\frac{1}{2})\), and the claim holds with \(x_{n,0}=0\in [ -\frac{1}{n-1},0 ] \).
If \(k=n-1\), then we have \(\varphi _{n,n-1} ( x ) =\sum_{i=0} ^{n-2}\frac{1}{1-\frac{i}{n-1}x}>0\) for \(x\in [0,1)\), and the claim holds with \(x_{n,n-1}=1\in [ \frac{n-2}{n-1},1 ] \).
Assume now that \(n>2\) and \(k\in \{ 1,\ldots ,n-2 \} \). We will first show that if \(\varphi _{n,k} ( x ) =0\), then \(\varphi _{n,k}^{{\prime }} ( x ) <0\) (note that \(\varphi _{n,k} ( 0 ) =k>0\) and thus \(x\neq 0\)).
We have
$$\begin{aligned} \varphi _{n,k}^{{\prime }} ( x ) =&\sum _{i=0}^{n-1}\frac{ \frac{i}{n-1}}{ ( 1-\frac{i}{n-1}x ) ^{2}}-\sum _{i=0}^{n-k-1}\frac{1+ \frac{i}{n-1}}{ ( 1-x-\frac{i}{n-1}x ) ^{2}} \\ =&\frac{1}{x} \Biggl( \sum_{i=0}^{n-1} \frac{- ( 1-x\frac{i}{n-1} ) +1}{ ( 1-\frac{i}{n-1}x ) ^{2}}- \sum_{i=0}^{n-k-1} \frac{- ( 1-x ( 1+\frac{i}{n-1} ) ) +1}{ ( 1-x-\frac{i}{n-1}x ) ^{2}} \Biggr) \\ =&-\frac{1}{x}\varphi _{n,k} ( x ) +\frac{1}{x} \Biggl( \sum_{i=0}^{n-1}\frac{1}{ ( 1-\frac{i}{n-1}x ) ^{2}}-\sum _{i=0}^{n-k-1}\frac{1}{ ( 1-x-\frac{i}{n-1}x ) ^{2}} \Biggr) . \end{aligned}$$
If
\(\varphi _{n,k} ( x ) =0\), then we obtain
\(\varphi _{n,k} ^{{\prime }} ( x ) =\frac{1}{x} ( \sum_{i=0}^{n-1}\frac{1}{ ( 1-\frac{i}{n-1}x ) ^{2}}-\sum_{i=0}^{n-k-1}\frac{1}{ ( 1-x-\frac{i}{n-1}x ) ^{2}} ) \), and we are left to prove the implication
$$ \sum_{i=0}^{n-1}\frac{1}{1-\frac{i}{n-1}x}=\sum _{i=0}^{n-k-1}\frac{1}{1-x- \frac{i}{n-1}x}\quad \Rightarrow\quad \sum _{i=0}^{n-1}\frac{1}{ ( 1- \frac{i}{n-1}x ) ^{2}}< \sum _{i=0}^{n-k-1}\frac{1}{ ( 1-x- \frac{i}{n-1}x ) ^{2}}. $$
Choosing
\(a_{i+1}=\frac{1}{1-\frac{i}{n-1}x}\),
\(i\in \{ 0, \ldots ,n-1 \} \), and
\(b_{j+1}=\frac{1}{1-x-\frac{j}{n-1}x}\),
\(j\in \{ 0,\ldots ,n-k-1 \} \), we have
\(\max_{1\leq i \leq n}a_{i}=a_{n}=\frac{1}{1-x}=b_{1}=\min_{1\leq j\leq n-k}b_{j}\), and the implication follows from Lemma
1, concluding the proof of the claim.
We showed that \(\varphi _{n,k} ( x ) =0\) implies \(\varphi _{n,k} ^{{\prime }} ( x ) <0\). Since \(\varphi _{n,k}\) is continuously differentiable, a moment’s thought shows that this condition implies that \(\varphi _{n,k}\) can change signs at most once on the interval \([0,\frac{n-1}{2n-k-2} )\).
Since \(\varphi _{n,k} ( 0 ) =n- ( n-k ) =k>0\) and \(\lim_{x\nearrow \frac{n-1}{2n-k-2}}\varphi _{n,k} ( x ) =- \infty \), the function \(\varphi _{n,k}\) changes sign on \([0, \frac{n-1}{2n-k-2} )\); let \(x_{n,k}\) denote its unique root. We are left to show that \(x_{n,k}\) belongs to the specified interval.
Using Lemma
3 with
\(N=n-1\) and
\(f ( t ) =\frac{1}{1-tx}\), respectively, with
\(N=n-k-1\) and
\(f ( t ) =\frac{1}{1-x-t\frac{n-k-1}{n-1}x}\), we obtain:
$$\begin{aligned} \varphi _{n,k} ( x ) &\leq \biggl( ( n-1 ) \int _{0}^{1}\frac{1}{1-tx}\,dt+ \frac{1+\frac{1}{1-x}}{2}+\frac{\frac{x}{ ( 1-x ) ^{2}}-x}{4 ( n-1 ) } \biggr) \\ &\quad {}- \biggl( ( n-k-1 ) \int _{0}^{1}\frac{1}{1-x-tx \frac{n-k-1}{n-1}}\,dt+ \frac{\frac{1}{1-x}+\frac{1}{1-x- \frac{n-k-1}{n-1}x}}{2} \biggr) \\ & = \biggl( -\frac{n-1}{x}\ln ( 1-x ) +\frac{1+ \frac{1}{1-x}}{2}+ \frac{\frac{x}{ ( 1-x ) ^{2}}-x}{4 ( n-1 ) } \biggr) \\ &\quad {}- \biggl( -\frac{n-1}{x}\ln \frac{1-x-\frac{n-k-1}{n-1}x}{1-x}+\frac{ \frac{1}{1-x}+\frac{1}{1-x-\frac{n-k-1}{n-1}x}}{2} \biggr) \\ & =\frac{1}{2} \biggl( 1-\frac{1}{1-x-\frac{n-k-1}{n-1}x} \biggr) + \frac{x ^{2} ( 2-x ) }{4 ( n-1 ) ( 1-x ) ^{2}}+\frac{n-1}{x}\ln \frac{1-x-\frac{n-k-1}{n-1}x}{ ( 1-x ) ^{2}}. \end{aligned}$$
In particular, for \(x=\frac{k}{n-1}\), we obtain \(\varphi _{n,k} ( \frac{k}{n-1} ) \leq \frac{k ( 2n-k-2 ) ( k-2 ( n-1 ) ^{2} ) }{4 ( n-1 ) ^{2} ( n-k-1 ) ^{2}}<0\), which shows that \(x_{n,k}<\frac{k}{n-1}\).
To obtain the lower bound for \(x_{n,k}\), first note that for \(k=1\), the claim is trivial (\(\varphi _{n,1} ( 0 ) =1>0\), and thus \(x_{n,1}>0\)), so we may assume that \(k\in \{ 2,\ldots ,n-1 \} \).
Using again Lemma
3 with the same choices as before, we obtain
$$\begin{aligned} \varphi _{n,k} ( x ) \geq & \biggl( ( n-1 ) \int _{0}^{1}\frac{1}{1-tx}\,dt+ \frac{1+\frac{1}{1-x}}{2} \biggr) - \biggl( ( n-k-1 ) \int _{0}^{1}\frac{1}{1-x-tx\frac{n-k-1}{n-1}}\,dt \\ &{}+\frac{\frac{1}{1-x}+\frac{1}{1-x-\frac{n-k-1}{n-1}x}}{2}+\frac{\frac{ \frac{n-k-1}{n-1}x}{ ( 1-x-\frac{n-k-1}{n-1}x ) ^{2}}-\frac{ \frac{n-k-1}{n-1}x}{ ( 1-x ) ^{2}}}{4 ( n-k-1 ) } \biggr) \\ =& \biggl( -\frac{n-1}{x}\ln ( 1-x ) +\frac{1+ \frac{1}{1-x}}{2} \biggr) - \biggl( -\frac{n-1}{x}\ln \frac{1-x- \frac{n-k-1}{n-1}x}{1-x} \\ &{} +\frac{\frac{1}{1-x}+\frac{1}{1-x-\frac{n-k-1}{n-1}x}}{2}+\frac{\frac{ \frac{n-k-1}{n-1}x}{ ( 1-x-\frac{n-k-1}{n-1}x ) ^{2}}-\frac{ \frac{n-k-1}{n-1}x}{ ( 1-x ) ^{2}}}{4 ( n-k-1 ) } \biggr) \\ =&\frac{1}{2} \biggl( 1-\frac{1}{1-x-\frac{n-k-1}{n-1}x} \biggr) - \biggl( \frac{1}{ ( 1-x-\frac{n-k-1}{n-1}x ) ^{2}}-\frac{1}{ ( 1-x ) ^{2}} \biggr) \frac{x}{4 ( n-1 ) } \\ &{} +\frac{n-1}{x}\ln \frac{1-x-\frac{n-k-1}{n-1}x}{ ( 1-x ) ^{2}}. \end{aligned}$$
To simplify the following computation, denote
\(A= ( \frac{n-k}{n-1} ) ^{2}\),
\(B= \frac{k-1}{ ( n-1 ) ^{2}}\), and
\(C=\frac{A}{B}=\frac{ ( n-k ) ^{2}}{k-1}\). For
\(x=\frac{k-1}{n-1}\), the inequality becomes
$$\begin{aligned} \varphi _{n,k} \biggl( \frac{k-1}{n-1} \biggr) \geq & \frac{1}{2} \biggl( 1-\frac{1}{A+B} \biggr) - \biggl( \frac{1}{ ( A+B ) ^{2}}-\frac{1}{A ^{2}} \biggr) \frac{B}{4}+ \frac{1}{B}\ln \biggl( \frac{A+B}{A} \biggr) \\ = & \frac{1}{2}+ \frac{1}{B} \biggl(-\frac{1}{2 ( 1+C ) }- \frac{1}{4 ( 1+C ) ^{2}}+\frac{1}{4C^{2}}+\ln \biggl( 1+\frac{1}{C} \biggr) \biggr). \end{aligned}$$
Since
$$\begin{aligned} &\frac{d}{dC} \biggl( -\frac{1}{2 ( 1+C ) }-\frac{1}{4 ( 1+C ) ^{2}}+ \frac{1}{4C^{2}}+\ln \biggl( 1+\frac{1}{C} \biggr) \biggr) \\ &\quad = \biggl( \frac{1}{2 ( 1+C ) ^{2}}-\frac{1}{C ( 1+C ) } \biggr) + \biggl( \frac{1}{2 ( 1+C ) ^{3}}-\frac{1}{2C ^{3}} \biggr) < 0 \end{aligned}$$
and
\(\lim_{C\rightarrow \infty } ( -\frac{1}{2 ( 1+C ) }-\frac{1}{4 ( 1+C ) ^{2}}+\frac{1}{4C^{2}}+\ln ( 1+ \frac{1}{C} ) ) =0\), we conclude that
\(( -\frac{1}{2 ( 1+C ) }-\frac{1}{4 ( 1+C ) ^{2}}+\frac{1}{4C ^{2}}+\ln ( 1+\frac{1}{C} ) ) >0\) for all
\(C>0\).
From this and from the previous inequality we obtain \(x_{n,k}> \frac{k-1}{n-1}\), concluding the proof. □