1. Introduction
The
binomial coefficients are defined, for non-negative integers \(n\) and \(m\), by \[\binom{n}{m} =
\begin{cases}
\dfrac{n!}{m!(n – m)!}, & \text{if } n \geq m;\\
0, & \text{if } n < m.
\end{cases}\] More generally, for complex numbers \(r\) and \(s\), they are defined by \[\binom{r}{s} = \frac{\Gamma(r+1)}{\Gamma(s+1)
\Gamma(r-s+1)},\] where the Gamma function, \(\Gamma(z)\), is defined for \(\Re(z) > 0\) by the integral [1] \[\Gamma(z) = \int_0^\infty e^{-t} t^{z – 1} \,
dt.\] The Gamma function can be extended to the entire complex
plane through analytic continuation. It possesses simple poles at each
of the points \(z = \ldots, -3, -2, -1,
0\). The Gamma function extends the classical factorial function
to the complex plane via the relation \((z-1)!
= \Gamma(z)\), thereby facilitating the computation of binomial
coefficients for non-integer and non-real values.
Closely related to the Gamma function is the psi or digamma function,
defined by \(\psi(z) =
\Gamma'(z)/\Gamma(z)\). It has the infinite series
representation [1, p.14]
\[\label{psi_expression}
\psi(z) = -\gamma + \sum_{k=0}^\infty \left( \frac{1}{k+1} –
\frac{1}{k+z} \right),\tag{1}\] where \(\gamma\) is the Euler-Mascheroni
constant.
Harmonic numbers, denoted by \(H_x\)
for \(x \in \mathbb{C} \setminus
\mathbb{Z}^-\), are defined by the recurrence relation \[\label{recurrence}
H_x = H_{x – 1} + \frac{1}{x}, \quad H_0 = 0.\tag{2}\] They are related
to the digamma function through the fundamental relation \[\label{eq.cx26aju}
H_x = \psi(x + 1) + \gamma.\tag{3}\] When \(x\) is a positive integer, say \(n\), the harmonic numbers form the sequence
\((H_n)_{n \in \mathbb{Z}^+}\), and the
recurrence relation (2) yields \[H_n = \sum_{k=1}^n \frac{1}{k}, \quad H_0 =
0.\tag{4}\]
The following combinatorial identity is attributed to Frisch [2]: \[\label{Id_Frisch}
\sum_{k=0}^n (-1)^k \frac{\binom{n}{k}}{\binom{b+k}{c}} = \frac{c}{n +
c} \cdot \frac{1}{\binom{n + b}{b – c}}, \quad \text{for } n \in
\mathbb{Z}^+ \text{ and } b,\, c,\, b – c \in \mathbb{C} \setminus
\mathbb{Z}^-.\tag{5}\] This identity is listed in Gould’s compendium
[3] as Identity 4.2 and
was recently utilized by Gould and Quaintance [4] to establish a new binomial transform
identity. Abel [5] provided
a concise proof of Frisch’s identity and investigated its infinite
variant.
In this note, we demonstrate how Frisch’s identity (5) can be applied to prove and
generalize several well-known identities involving harmonic numbers
\(H_n\). Additionally, we present some
combinatorial identities involving odd harmonic numbers \(O_n\), which can be readily derived from
our results. Our approach primarily involves leveraging the fact that
derivatives of generalized binomial coefficients yield harmonic numbers.
This method is well-established and has been employed in significant
earlier research by other mathematicians [6-9].
2. Results
Theorem 1. For \(n\in\mathbb Z^+\) and \(b,\, c,\, b – c\in\mathbb C\setminus\mathbb
Z^{-}\), we have \[\label{main_id1}
\sum_{k=0}^n (-1)^k \frac{\binom{n}{k}}{\binom{b+k}{c}}\left ( H_{k+b} –
H_{k+b-c}\right )
= \frac{c}{n+c} \frac{H_{n+b}-H_{b-c}}{\binom{n+b}{n+c}}\tag{6}\] and
\[\label{main_id1b}
\sum_{k = 0}^n (- 1)^k \binom{{n}}{k}\frac{c}{{k + c}}\frac{{H_{k + b} –
H_{b – c}}}{{\binom{{k + b}}{{k + c}}}}
= \frac{{H_{n + b} – H_{n + b – c} }}{{\binom{{b + n}}{c}}}.\tag{7}\] In
particular, for \(n\in\mathbb Z^+\) and
\(b\in\mathbb C\setminus\mathbb Z^{-}\)
we have \[\label{main_id11}
\sum_{k=0}^n (-1)^k \frac{\binom {n}{k}}{\binom{b+k}{k}}\left ( H_{k+b}
– H_{k}\right )
= \frac{b}{n+b} H_{n+b}\tag{8}\] and \[\label{main_id11b}
\sum_{k = 0}^n (- 1)^k \binom{{n}}{k}\frac{{H_{k + b} }}{{k + b}} =
\frac{{H_{n + b} – H_n }}{{b\binom{{b + n}}{b}}}.\tag{9}\]
Proof. In Frisch’s identity (5)
treat \(b\) and \(c\) as complex numbers and differentiate
w.r.t. \(b\), using \[\label{eq.u72pdmn}
\frac{d}{db} \binom{b+k}{c}^{-1} = \binom{b+k}{c}^{-1}\left
(\psi(b+k+1-c)-\psi(b+k+1)\right )\tag{10}\] and \[\label{eq.lilpf1k}
\frac{d}{db} \binom{n+b}{n+c}^{-1} = \binom{n+b}{n+c}^{-1}\left
(\psi(b+1-c)-\psi(b+n+1)\right ),\tag{11}\] simplify, making use of the
fundamental relation (3). This gives (6). Identity (7)
is the binomial transform of (6). ◻
Corollary 2. For \(n\in\mathbb Z^+\) and \(b\in\mathbb C\setminus\mathbb Z^{-}\),
\(b\ne 0\) we have \[\label{cor_id1}
\sum_{k=0}^n (-1)^k \frac{\binom{n}{k}}{(b+k)^2} = \frac{1}{n+1}
\frac{H_{n+b}-H_{b-1}}{\binom{n+b}{n+1}}.\tag{12}\]
Identity (12) generalizes the well-known
identity \[\sum_{k=0}^n (-1)^k
\frac{\binom{n}{k}}{(k+1)^2} = \frac{H_{n+1}}{n+1},\tag{13}\] which can
be proved directly via \[\begin{aligned}
(n+1) \sum_{k=0}^n \binom {n}{k} \frac{(-1)^k}{(k+1)^2} &=
\sum_{k=0}^n \frac{n+1}{k+1} \binom {n}{k} \frac{(-1)^k}{k+1} \\
&= \sum_{k=0}^n \binom {n+1}{k+1} \frac{(-1)^k}{k+1} \\
&= \sum_{k=1}^{n+1} \binom {n+1}{k} \frac{(-1)^{k-1}}{k} \\
&= H_{n+1}.
\end{aligned}\]
Theorem 3. For \(n\in\mathbb Z^+\) and \(b,\, c,\, b – c\in\mathbb C\setminus\mathbb
Z^{-}\) we have \[\label{main_id2}
\sum_{k=0}^n (-1)^{k+1} \frac{\binom{n}{k}}{\binom{b+k}{c}}\left (
H_{k+b-c} – H_{c}\right )
= \frac{1}{(n+c)\binom{n+b}{n+c}} \left (\frac{n}{n+c} + c
(H_{n+c}-H_{b-c})\right ).\tag{14}\] In particular, for \(n\in\mathbb Z^+\) and \(b\in\mathbb C\setminus\mathbb Z^{-}\) we
have \[\label{main_id22}
\sum_{k=0}^n (-1)^{k+1} \binom{n}{k} H_{k+b} =
\frac{1}{n\binom{n+b}{n}}.\tag{15}\]
Proof. Differentiate Frisch’s identity (5)
w.r.t. \(c\) using \[\frac{d}{dc} \binom{b+k}{c}^{-1} =
\binom{b+k}{c}^{-1}\left (\psi(c+1)-\psi(b+k+1-c)\right ),\] and
the proof is completed. ◻
Identity (15) generalizes the well-known
identity \[\sum_{k=0}^n (-1)^{k} \binom{n}{k}
H_{k} = – \frac{1}{n},\tag{16}\] which is the binomial transform of the
sequence \(H_n\) (see [10]).
Corollary 4. For \(n\in\mathbb N_0\) and \(b\in\mathbb C\setminus\mathbb Z^{-}\) we
have \[\label{eq.auhzvb5}
\sum_{k=1}^n (-1)^{k+1} \frac{\binom{n}{k}}{k\binom{k+b}{k}} = H_{n+b} –
H_{b}.\tag{17}\] In particular, \[\sum_{k=1}^n
(-1)^{k+1} \frac{\binom{n}{k}}{k\binom{k+n}{k}} = H_{2n} –
H_{n}.\tag{18}\]
Corollary 5. For \(n\in\mathbb Z^+\) and \(b\in\mathbb C\setminus\mathbb Z^{-}\) we
have \[\label{cor_id2}
\sum_{k=0}^n (-1)^k \frac{\binom{n}{k}}{\binom{k+b}{k}} H_{k+b} =
\frac{b}{n+b} H_{b} – \frac{n}{(n+b)^2}.\tag{19}\]
Proof. Set \(c=b\) in (14) to get \[\sum_{k=0}^n (-1)^{k+1}
\frac{\binom{n}{k}}{\binom{b+k}{b}}\left ( H_{k} – H_{b}\right )
= \frac{1}{n+b} \left (\frac{n}{n+b} + b H_{n+b}\right ).\] Use
the fact that \[\label{eq.twi4acl}
\sum_{k=0}^n (-1)^{k} \frac{\binom{n}{k}}{\binom{b+k}{k}} =
\frac{b}{n+b}\tag{20}\] and combine with (8). ◻
Corollary 6. For \(n\in\mathbb N_0\) and \(b\in\mathbb C\setminus\mathbb Z^{-}\) we
have \[\label{cor_id3}
\sum_{k=1}^n \binom{n}{k} (-1)^{k+1} \frac{k}{(k+b)^2} = \frac{H_{n+b} –
H_{b}}{\binom{n+b}{n}}.\tag{21}\] In particular, \[\sum_{k=1}^n \binom{n}{k} (-1)^{k+1}
\frac{k}{(k+n)^2} = \frac{H_{2n} –
H_{n}}{\binom{2n}{n}}.\tag{22}\]
Proof. Using the fact that the right hand-side of (19) is the binomial transform of
\(H_{n+b}/\binom{n+b}{n}\) we get \[\sum_{k=0}^n \binom{n}{k} (-1)^k \left (
\frac{b}{k+b} H_{b} – \frac{k}{(k+b)^2}\right ) =
\frac{H_{n+b}}{\binom{n+b}{n}}.\] Combine this with \[\label{binomial_frac_id}
\sum_{k=0}^n \binom{n}{k} (-1)^k \frac{1}{k+b} = \frac{1}{b
\binom{n+b}{b}}\tag{23}\] and the proof is completed. ◻
Theorem 7. If \(n\in\mathbb Z^+\) and \(b\in\mathbb C\setminus\mathbb Z^{-}\), then
\[\label{eq.hzbwh6o}
\sum_{k = 0}^n {( – 1)^{k – 1} \frac{{\binom{{n}}{k}}}{{\binom{{b +
k}}{k}}}H_k }
= \frac{b}{{n + b}}\left( {H_{n + b} – H_b } \right) + \frac{n}{{\left(
{n + b} \right)^2 }}.\tag{24}\]
Proof. Set \(c=b\) in (14) and use (20). ◻
Theorem 8. If \(n\in\mathbb N_0\) and \(b\in\mathbb C\setminus\mathbb Z^{-}\),
\(b\ne 0\), then \[\label{eq.ufus43q}
\sum_{k = 0}^n \binom{{n}}{k} \frac{(- 1)^k}{(k + b)^3} = \frac{1}{2n +
2}\binom{{n + b}}{{n + 1}}^{- 1} \left( {\left( {H_{n + b}
– H_{b – 1} } \right)^2 – H_{b – 1}^{(2)} + H_{n + b}^{(2)} }
\right).\tag{25}\] In particular, \[\label{Bai_id}
\sum_{k = 0}^n \binom{{n}}{k} \frac{(- 1)^k}{(k + 1)^3} = \frac{1}{2n +
2}\left( {H_{n + 1}^2 + H_{n + 1}^{(2)} } \right).\tag{26}\]
Proof. Write (12) as \[\sum_{k=0}^n (-1)^k \frac{\binom{n}{k}}{(b+k)^2}
= \frac{1}{n+1} \frac{\psi(n + b + 1)-\psi(b)}{\binom{n + b}{n +
1}}\] and differentiate with respect to \(b\) to obtain \[\begin{split}
&\sum_{k = 0}^n {\frac{{( – 1)^k }}{{(k + b)^3 }}\binom{{n}}{k}} \\
&\qquad = \frac{1}{2n + 2} \binom{{n + b}}{{n + 1}}^{ – 1} \left(
{\psi (n + b + 1) – \psi (b)} \right)^2 \\
&\qquad\quad – \frac{1}{{2n + 2}}\binom{{n + b}}{{n + 1}}^{ – 1}
\left( {\psi _1 (n + b + 1) – \psi _1 (b)} \right),
\end{split}\] where \(\psi _1
(x)\) is the trigamma function defined by \[\psi _1 (x) = \frac{d}{{dx}}\psi (x) = \sum_{k =
0}^\infty \frac{1}{(k + x)^2}\] and \(H^{(2)}_r\) is the \(r^{\,th}\) second order harmonic number,
\[H_r^{(2)} = \sum_{j = 1}^r
\frac{1}{j^2}.\] The result follows upon using the fact that
\[\label{eq.kmcmfjf}
\psi \left( {x + 1} \right) – \psi \left( {y + 1} \right) = H_x –
H_y\tag{27}\] and \[\label{eq.ezvh9zc}
\psi_1 \left( {x + 1} \right) – \psi_1 \left( {y + 1} \right) =
H_y^{(2)} – H_x^{(2)},\tag{28}\] for \(x,y\in\mathbb C\setminus\mathbb
Z^{-}\). ◻
Theorem 9. If \(n\in\mathbb N_0\) and \(b\in\mathbb C\setminus\mathbb Z^{-}\), then
\[
\sum_{k = 0}^n \left( { – 1} \right)^{k – 1} \binom{{n}}{k}H_{k +
b}^{(2)} = \frac{{H_{n + b} – H_b }}{{n\binom{{n + b}}{n}}},\quad n\ne
0, \tag{29}\]\[
\sum_{k = 1}^n \left( { – 1} \right)^{k – 1} \binom{{n}}{k}\frac{{H_{k +
b} – H_b }}{{k\binom{{k + b}}{k}}} = H_{n + b}^{(2)} –
H_b^{(2)}, \tag{30}\]\[
\sum_{k = 1}^n \left( { – 1} \right)^{k – 1} \binom{{n}}{k}\frac{{H_{k +
b} }}{{k\binom{{k + b}}{k}}}
= \left( {H_{n + b} – H_b } \right)H_b + H_{n + b}^{(2)} –
H_b^{(2)}. \tag{31}\]
Proof. Differentiate (15)
with respect to \(b\) to get (29). Identity (30) is the inverse of (29). Identity (31) is obtained by using (17) in (30). ◻
Theorem 10. If \(n\in\mathbb Z^+\) and \(b\in\mathbb C\setminus\mathbb Z^{-}\), then
\[\label{eq.iauweap}
\begin{split}
\sum_{k = 0}^n {( – 1)^{k – 1} \frac{{\binom{{n}}{k}}}{{\binom{{b +
k}}{b}}}H_k H_{k + b} }
&= \frac{1}{{n + b}}\left( {bH_b – \frac{n}{{n + b}}} \right)\left(
{H_{n + b} – H_b } \right)\\
&\qquad + \frac{b}{{n + b}}\left( {H_{n + b}^{(2)} – H_b^{(2)} }
\right) + \frac{n}{{\left( {n + b} \right)^2 }}\left( {\frac{2}{{n + b}}
+ H_b } \right).
\end{split}\tag{32}\]
Proof. Write (24)
as \[\sum_{k = 0}^n {( – 1)^{k – 1}
\frac{{\binom{{n}}{k}}}{{\binom{{b + k}}{k}}}H_k }
= \frac{b}{{n + b}}\left( {\psi(n + b + 1) – \psi(b + 1) } \right) +
\frac{n}{{\left( {n + b} \right)^2 }}\] and differentiate with
respect to \(b\), using (10). Use (24)
again in simplifying the left hand side of the resulting
expression. ◻
In particular, for all positive integers \(n\), we have \[\label{id_Hk_squared}
\sum_{k = 0}^n (- 1)^k \binom{n}{k} H_k^2 = \frac{{H_n }}{n} –
\frac{2}{n^2}\tag{33}\] and \[\begin{aligned}
&\sum_{k = 0}^n (- 1)^{k – 1} \frac{\binom{n}{k}}{\binom{{n +
k}}{k}} H_k H_{k + n} \nonumber \\
&\qquad = \frac{1}{2}\left( H_{2n}^{(2)} – H_n^{(2)} \right) +
\frac{1}{2} H_{2n} \left( H_{n} – \frac{1}{2n} \right)
– \frac{1}{2} H_{n} H_{n-1} + \frac{1}{4n^2}.
\end{aligned}\tag{34}\]
Corollary 11. If \(n\) is a positive integer, then \[\label{eq.vsafq0i}
\sum_{k=1}^n (- 1)^{k – 1} \binom{n}{k} \frac{H_k}{k} =
H_n^{(2)}.\tag{35}\]
Proof. Treat identity (33) as the binomial
transform of the sequence \(H_n^2\).
The inverse relation yields \[\sum_{k=1}^n (-
1)^{k} \binom{n}{k} \frac{H_k}{k} + 2 \sum_{k=1}^n \binom{n}{k}
\frac{(-1)^{k-1}}{k^2} = H_n^{2}.\] But (see Boyadzhiev’s book
[10, p.64]) \[\sum_{k=1}^n \binom{n}{k} \frac{(-1)^{k-1}}{k^2}
= \frac{1}{2}\left ( H_n^2 + H_n^{(2)}\right )\] and the proof is
completed. ◻
3. Some Identities
Involving Odd Harmonic Numbers
This section contains some combinatorial identities involving odd
harmonic numbers \(O_n\), which are
defined by \[O_n = \sum_{k = 1}^n
\frac{1}{2k-1},\quad O_0=0.\] Obvious relations between harmonic
numbers \(H_n\) and odd harmonic
numbers \(O_n\) are given by \[\label{eq.h9wjxs0}
H_{2n} = \frac{1}{2} H_n + O_n \qquad \text{and} \qquad H_{2n – 1} =
\frac{1}{2}H_{n – 1} + O_n.\tag{36}\] Additional relations are contained
in the next lemma.
Lemma 1. If \(n\) is an integer, then \[
H_{n – 1/2} =2O_n – 2\ln 2,\tag{37}\]\[
H_{n – 1/2} – H_{ – 1/2} = 2O_n,\tag{38}\]\[
H_{n – 1/2} – H_{1/2} = 2\left( {O_n – 1} \right),\tag{39}\]\[
H_{n + 1/2} – H_{ – 1/2} = 2O_{n + 1},\tag{40}\]\[
H_{n + 1/2} – H_{1/2} = 2\left( {O_{n + 1} – 1} \right),\tag{41}\]\[
H_{n + 1/2} – H_{n – 1/2} = \frac{2}{{2n + 1}},\tag{42}\]\[
H_{n – 1/2} – H_{-3/2} = 2\left( {O_n – 1} \right), \tag{43}\]\[
H_{n + 1/2} – H_{-3/2} = 2\left( {O_{n + 1} – 1} \right).\tag{44}\]
Proof. Use (3) as the definition of the
harmonic numbers for all complex \(n\)
(excluding zero and the negative integers) and use the known result for
the digamma function at half-integer arguments [1, Eq. (51)], namely, \[\psi (n + 1/2) = – \gamma – 2\ln 2 + 2\sum_{k =
1}^n \frac{1}{2k – 1}.\] ◻
Lemma 2 (Gould [3]). If \(r\) and \(s\) are integers such that \(0\le s\le r\), then \[\label{eq.cwrdtmu}
\binom{{r + 1/2}}{s} = \binom{{2r + 1}}{{2s}}\binom{{r}}{s}^{ – 1} 2^{ –
2s} \binom{{2s}}{s}\tag{45}\] and \[\label{eq.s0t6h30}
\binom{{r – 1/2}}{s} = \binom{{r}}{s}\binom{{2r – 2s}}{{r – s}}^{ – 1}
2^{ – 2s} \binom{{2r}}{r}.\tag{46}\] We also have \[
\binom{{r}}{{1/2}} = \dfrac{{2^{2r + 1} }}{{\pi
\binom{{2r}}{r}}},\quad\text{[3], Identity Z.48}, \tag{47}\]\[
\binom{{r}}{{ – 1/2}} = \dfrac{{2^{2r + 1} }}{{\pi (2r +
1)\binom{{2r}}{r}}}\tag{48}\] and \[
\binom{{r – 1/2}}{{r + 1}} = – \frac{1}{{r +
1}}\binom{{2r}}{r}\frac{1}{{2^{2r + 1} }}.\tag{49}\]
Proof. Identity (48)
follows from the fact that \[\binom{{r}}{{ –
1/2}} = \frac{{r!}}{{( – 1/2)!(r + 1/2)!}},\] while (49) is a consequence of \[\binom{{r – 1/2}}{{r + 1}} = –
\frac{1}{2}\frac{{(r – 1/2)!}}{{(r + 1)!\sqrt \pi }}.\tag{50}\] ◻
Theorem 12. If \(n\) is a non-negative integer, then \[
\sum_{k = 0}^n \binom{n}{k} \frac{(- 1)^k}{\left( {2k + 1} \right)^2}
= \frac{2^{2n + 1}}{n+1} \frac{O_{n+1}}{\binom{2(n + 1)}{n + 1}},
\tag{51}\]\[
\sum_{k = 0}^n \binom{n}{k} (- 1)^k \frac{2^{2k + 1}}{k+1}
\frac{O_{k+1}}{\binom{2(k + 1)}{k + 1}} = \frac{1}{(2n+1)^2},\tag{52}\]\[
\sum_{k = 0}^n \binom{{n}}{k} \frac{(- 1)^{k – 1}}{\left( {2k – 1}
\right)^2} = 2^{2n} \frac{{O_n – 1}}{\binom{{2n}}{n}}
, \tag{53}\] and \[\label{eq.fhj2wwz}
\sum_{k = 0}^n \binom{{n}}{k}(- 1)^{k – 1} 2^{2k}
\frac{O_k}{\binom{{2k}}{k}} = \frac{2n}{(2n – 1)^2}.\tag{54}\]
Proof. The first identity is obtained by setting \(b=1/2\) in (12) and
using (40) and (45); while the second is its
inverse transform. Identity (53)
is obtained by setting \(b=-1/2\) in (12) and using (43)
and (49). To prove identity (54) use the inverse binomial
transform of (53) to get \[\sum_{k = 0}^n \binom{{n}}{k}(- 1)^{k – 1} 2^{2k}
\frac{O_k}{\binom{{2k}}{k}}
– \sum_{k = 0}^n \binom{{n}}{k}(- 1)^{k – 1}
\frac{2^{2k}}{\binom{{2k}}{k}} = \frac{1}{(2n-1)^2}.\] But \[\label{eq.xyz1}
\sum_{k = 0}^n \binom{{n}}{k}(- 1)^{k – 1}
\frac{2^{2k}}{\binom{{2k}}{k}} = \frac{1}{2n-1}\tag{55}\] and the proof
of (54) is completed. ◻
Theorem 13. If \(n\) is a non-negative integer, then \[
\sum_{k = 0}^n {( – 1)^{k + 1} \binom{{n}}{k}\frac{{2k – 1}}{{2^{2(k –
1)} }}\binom{{2(k – 1)}}{{k – 1}}O_k } = \frac{1}{{2^{2(n – 1)}
}}\binom{{2(n – 1)}}{{n – 1}}\left( {\frac{{2n}}{{2n – 1}} – O_n }
\right), \tag{56}\]\[
\sum_{k = 0}^n {( – 1)^{k + 1} \binom{{n}}{k}\frac{{2k + 1}}{{2^{2k}
}}\binom{{2k}}{k}O_{k + 1} } = \frac{1}{{2n –
1}}\frac{{\binom{{2n}}{n}}}{{2^{2n} }}\left( {\frac{{4n – 1}}{{2n – 1}}
– O_n } \right), \tag{57}\]\[
\sum_{k = 0}^n {( – 1)^{k + 1} \binom{{n}}{k}2^{ – 2k – 2} \binom{{2(k +
1)}}{{k + 1}}(O_{k + 1} – 1) }
= \frac{2^{ – 2n – 2}}{{2n + 1}} \binom{{2(n + 1)}}{{n + 1}}\left( {O_{n
+ 1} -\frac{1}{{2n + 1}}} \right), \tag{58}\]\[
\sum_{k = 0}^n {( – 1)^{k + 1} \binom{{n}}{k}2^{ – 2k}
\binom{{2k}}{k}(O_k – 1) }
= 2^{ – 2n} \binom{{2n}}{n}\left( {\frac{{2n}}{{2n + 1}} + O_{n + 1}}
\right)\tag{59}\]
Proof. These results follow from (14). To
obtain (56), set \(b=-1\), \(c=-1/2\) in (14) and
use (38) and (48). Identity (57) comes from setting \(b=0\), \(c=-1/2\) in (14) and
using (40), (47)
and (48). ◻
Theorem 14. If \(n\) is a positive integer, then \[
\sum_{k = 1}^n \binom{{n}}{k} ( – 1)^{k + 1} O_k = \frac{{2^{2n – 1}
}}{{n \binom{{2n}}{n}}}, \tag{60}\]\[
\sum_{k = 1}^n \binom{{n}}{k} ( – 1)^{k + 1} \frac{2^{2k-1}}{k
\binom{{2k}}{k}} = O_n , \tag{61}\]\[
\sum_{k = 0}^n \binom{{n}}{k} ( – 1)^{k + 1} O_{k + 1} = \frac{{2^{2n –
1} }}{{n(2n + 1)\binom{{2n}}{n}}} , \tag{62}\]\[
\sum_{k = 1}^n \binom{{n}}{k} ( – 1)^{k + 1} \frac{{2^{2k – 1} }}{k(2k +
1) \binom{{2k}}{k}} = O_{n + 1} – 1 . \tag{63}\]
Proof. Write (15)
as \[\sum_{k = 0}^n {( – 1)^{k + 1}
\binom{{n}}{k}\left( {H_{k + b} – H_c } \right)} = \frac{1}{{n\binom{{n
+ b}}{n}}}.\] Evaluate at \((b,c)=(-1/2,-1/2)\) and at \((b,c)=(1/2,-1/2)\) to obtain (60) and (62). Identity (61) is the binomial transform
of (60) while (63)
is the transform of (62). ◻
Theorem 15. If \(n\) is a non-negative integer, then \[
\sum_{k = 1}^n \binom{{n}}{k}\frac{{(- 1)^{k + 1} k}}{(2k + 1)^2} =
\frac{{2^{2n – 1} (O_{n + 1} – 1)}}{(2n + 1)}, \tag{64}\]\[
\sum_{k = 1}^n \binom{{n}}{k}\frac{{(- 1)^{k + 1} k}}{(2k – 1)^2} =
\frac{{2^{2n – 1} O_n}}{\binom{{2n}}{n}}, \tag{65}\]\[
\sum_{k = 1}^n \binom{{n}}{k} ( – 1)^{k – 1} \frac{{2^{2k – 1} (O_{k +
1} – 1)}}{(2k + 1)\binom{{2k}}{k}} = \frac{n}{(2n + 1)^2}, \tag{66}\]\[
\sum_{k = 1}^n \binom{{n}}{k} ( – 1)^{k + 1}\frac{2^{2k – 1} O_k
}{\binom{{2k}}{k}} = \frac{n}{(2n – 1)^2} . \tag{67}\]
Proof. The first two results follow from (21). The
other two identities are the inverse binomial transforms of the
former. ◻
Theorem 16. If \(n\) is a non-negative integer, then \[
\sum_{k = 0}^n \binom{{n}}{k} (- 1)^{k – 1} \frac{2^{2k} H_k}{(2k+1)
\binom{{2k}}{k}}
= \frac{2O_{n + 1} }{2n + 1} – \frac{2}{(2n + 1)^2}, \tag{68}\]\[
\sum_{k = 0}^n \binom{{n}}{k} (- 1)^k \frac{2^{2k} H_k}{\binom{{2k}}{k}}
= \frac{2O_n}{{2n – 1}} – \frac{4n}{(2n – 1)^2}, \tag{69}\]\[
\sum_{k = 0}^n \binom{{n}}{k} (- 1)^k \frac{O_{k+1}}{2k+1}
= \frac{2^{2n+1}}{n+1} \frac{O_{n+1}}{\binom{2n+2}{n+1}} –
\frac{2^{2n-1}}{2n+1} \frac{H_n}{\binom{2n}{n}} , \tag{70}\]\[
\sum_{k = 0}^n \binom{{n}}{k} (- 1)^k \frac{O_{k}}{2k-1}
= \frac{2^{2n-1}}{\binom{2n}{n}} \left ( H_n – 2O_{n} \right ),
\tag{71}\]\[
\sum_{k = 1}^n {\left( { – 1} \right)^{k – 1}
\binom{{n}}{k}\frac{{2^{2k} }}{{\binom{{2k}}{k}}}H_{2k} } =
\frac{{4n}}{{\left( {2n – 1} \right)^2 }} – \frac{{O_n }}{{2n –
1}}.\tag{72}\]
Proof. The first two results follow from (24). Identity (70) is a consequence of the
inverse binomial relation of (68)
in conjunction with identity (51). Identity (71) is a consequence of the
inverse binomial relation of (69)
in conjunction with identity (65). Identity (72) is a consequence of (67) and (69)
on account of the first identity in (36). ◻
Theorem 17. If \(n\) is a non-negative integer, then \[\sum_{k = 0}^n {\binom{{n}}{k}\frac{{\left( { –
1} \right)^{k – 1} }}{{\left( {2k – 1} \right)^3 }}} = \dfrac{{2^{2n –
1} }}{{\binom{{2n}}{n}}}\left( {\left( {O_n – 1} \right)^2 +
O_n^{(2)} + 1} \right).\tag{73}\]
Proof. Set \(b=-1/2\) in (25) and use Lemma 46 and identity (47).
Use also \[\label{eq.cqhkqwq}
H_{n – 1/2}^{(2)} = – 2\zeta (2) + 4O_n^{(2)}\tag{74}\] which follows
from the known value \(H_{-1/2}^{(2)}=-2\zeta(2)\) since \[H_{n – 1/2}^{(2)} – H_{ – 1/2}^{(2)} =
4O_n^{(2)},\] where \(O_n^{(2)}\) is the \(n^{th}\) second order odd harmonic number
defined by \[O_n^{(2)} = \sum_{k = 1}^n
{\frac{1}{{\left( {2k – 1} \right)^2 }}}.\] ◻
Theorem 18. If \(n\) is a positive integer, then \[
\sum_{k = 1}^n {\left( { – 1} \right)^{k – 1} \binom{{n}}{k}O_k^{(2)}
} = \frac{{2^{2n – 1} }}{{\binom{{2n}}{n}}}\frac{{O_n
}}{n},\tag{75}\]\[
\sum_{k = 1}^n {\left( { – 1} \right)^{k – 1} \binom{{n}}{k}\frac{{2^{2k
– 1} }}{{\binom{{2k}}{k}}}\frac{{O_k }}{k}} =
O_n^{(2)}\tag{76}\]
Theorem 19. If \(n\) is a non-negative integer, then \[
\sum_{k = 0}^n {\left( { – 1} \right)^{k – 1}
\binom{{n}}{k}\frac{{2^{2k} }}{{\binom{{2k}}{k}}}H_k O_k } =
\frac{{8n}}{{\left( {2n – 1} \right)^3 }} – \frac{{4n\,O_n }}{{\left(
{2n – 1} \right)^2 }} – \frac{{2O_n^{(2)} }}{{\left( {2n – 1}
\right)}},\tag{77}\]\[
\sum_{k = 0}^n {\left( { – 1} \right)^k \binom{{n}}{k}\frac{{2^{2k}
}}{{\binom{{2k}}{k}}}H_k } = \frac{{2O_n }}{{\left( {2n – 1} \right)}}
– \frac{{4n}}{{\left( {2n – 1} \right)^2 }}. \tag{78}\]
Proof. Set \(b=-1/2\) in (32). Use Lemma 1, (46)
and (74). Equate rational
coefficients from both sides. Identity (78)
is a rediscovery of (69). ◻
In these notes, we have demonstrated how a combinatorial identity
attributed to Frisch can be used to prove a series of harmonic number
and odd harmonic number identities. Some of the results that we derived
are known and here we provide final comments and give further
references.
We begin by pointing out that identity (33) can also be found in
a recent article by Batir [11, Identity 18]. Next, identity (23) can be restated
as \[\sum_{k=0}^n \binom{n}{k} (-1)^k
\frac{b}{k+b} = \prod_{k=1}^n \frac{k}{b + k}.\] This identity is
known and two probabilistic proofs were given recently by Peterson [12] and Nakata [13]. Peterson [12] writes [cor_id1] in
the form \[\sum_{k=0}^n \binom{n}{k} (-1)^k
\left (\frac{b}{b+k}\right )^2 = \prod_{k=1}^n \frac{k}{b + k} \left (1
+ \sum_{k=1}^n \frac{b}{b+k}\right ),\] and also states an
expression for the generalization involving an additional parameter
\(m\) \[\label{Peterson}
\sum_{k=0}^n \binom{n}{k} (-1)^k \left (\frac{b}{b+k}\right )^m, \qquad
m\geq 1.\tag{79}\] In 2019, Bai and Luo [14] derived a new expression for Peterson’s
identity (79) using a partial fraction
decomposition and involving generalized harmonic numbers. As
applications of their main result, they stated some harmonic number
identities. One of their special cases is our identity (26).
It is remarkable that a simple expression for (79) is
also hidden in Frisch’s identity (5).
Indeed, setting \(c=1\) in (5) yields (23), i.e., \[\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{b+k} =
\frac{1}{b \binom{n+b}{b}}.\] Differentiating both sides \(m\) times w.r.t. \(b\) gives \[\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(b+k)^m}
= \frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{db^{m-1}}\frac{1}{b
\binom{n+b}{b}},\tag{80}\] which shows such a simple expression.
