Remarks on an Identity of Anastase and Díaz-Barrero

Abstract

We extend an algebraic identity of Anastase and Díaz-Barrero (2022) and apply our results to deduce various formulas for sums and series involving (among others) Fibonacci and Lucas numbers, Bernoulli polynomials, and the Riemann zeta function.

1. Introduction and Main Results

In an interesting paper published by Anastase and Díaz-Barrero [1] in 2022, the authors studied several inequalities involving the classical Fibonacci and Lucas numbers, $F_n$ and $L_n$. Among others, they proved that for $n\ge 2$,

$$\sum _{j=2}^n{\displaystyle \frac{F_j}{{(F_{j+3}-2)}^2}}\le {\displaystyle \frac{L_{n+1}+L_{n+3}-10}{20(F_{n+2}-1)}}.$$

In order to deduce their results, they applied certain discrete inequalities and the following algebraic identity.

Proposition 1.

Let $a_j$$(1\le j\le n;n\ge 3)$ be positive real numbers. Then,

$$\sum _{j=2}^n{a}_j{\left(\sum _{\nu =1}^{j-1}{a}_{\nu }\right)}^{-1}{\left(\sum _{\nu =1}^j{a}_{\nu }\right)}^{-1}=a_1^{-1}\sum _{j=2}^n{a}_j{\left(\sum _{j=1}^n{a}_j\right)}^{-1}.$$

(1)

Formula (1) (which also holds for $n=2$) can be written in a simpler form as

$$\sum _{j=2}^n{\displaystyle \frac{a_j}{A_{j-1}{A}_j}}={\displaystyle \frac{1}{a_1}}-{\displaystyle \frac{1}{A_n}},$$

(2)

where

$$A_{\nu }=\sum _{j=1}^{\nu }{a}_j.$$

The aim of this paper is to show that two modified versions of (2) can be used to obtain formulas for certain sums and series, like, for instance,

$${\displaystyle \frac{1}{2\sqrt{5}}}=1-\sum _{j=0}^{\infty }{\displaystyle \frac{F_{2^j}}{L_{2^j-1}{L}_{2^{j+1}-1}}}.$$

Anastase and Díaz-Barrero proved (1) by induction on n. Here, we apply a different method based upon the binomial theorem and telescoping summation to establish the following extensions of (2).

Theorem 1.

Let $a_j$$(1\le j\le n;n\ge 2)$ be non-zero complex numbers with $A_j\ne 0$$(1\le j\le n)$. Then, we have for integers $m\ge 0$ and $k\ge 1$,

$${\displaystyle \frac{1}{A_{m+n}^k}}={\displaystyle \frac{1}{A_{m+1}^k}}+\sum _{j=m+2}^{m+n}{\left(-{\displaystyle \frac{a_j}{A_{j-1}{A}_j}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left(-{\displaystyle \frac{A_j}{a_j}}\right)}^{\nu }$$

(3)

and

$${\displaystyle \frac{1}{A_{m+n}^k}}={\displaystyle \frac{1}{A_{m+1}^k}}-\sum _{j=m+2}^{m+n}{\displaystyle \frac{1}{A_j^k}}\sum _{\nu =1}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left({\displaystyle \frac{a_j}{A_{j-1}}}\right)}^{\nu }.$$

(4)

Proof. 

Let $2\le j\le n$, $m\ge 0$ and $k\ge 1$.

(i) Using the binomial theorem gives

$$A_{m+j-1}^k={(A_{m+j}-a_{m+j})}^k=\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right)A_{m+j}^{\nu }{(-a_{m+j})}^{k-\nu }+A_{m+j}^k.$$

Then

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{A_{m+n}^k}}-{\displaystyle \frac{1}{A_{m+1}^k}}& =\sum _{j=2}^n\left({\displaystyle \frac{1}{A_{m+j}^k}}-{\displaystyle \frac{1}{A_{m+j-1}^k}}\right)\hfill \\ \hfill & =\sum _{j=2}^n{\displaystyle \frac{A_{m+j-1}^k-A_{m+j}^k}{A_{m+j-1}^k{A}_{m+j}^k}}\hfill \\ \hfill & =\sum _{j=2}^n{\displaystyle \frac{1}{A_{m+j-1}^k{A}_{m+j}^k}}\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right)A_{m+j}^{\nu }{(-a_{m+j})}^{k-\nu }.\hfill \end{array}$$

This leads to (3).

(ii) We have

$$A_{m+j}^k={(A_{m+j-1}+a_{m+j})}^k=A_{m+j-1}^k+\sum _{\nu =1}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right)A_{m+j-1}^{k-\nu }{a}_{m+j}^{\nu }.$$

It follows that

$${\displaystyle \frac{1}{A_{m+n}^k}}-{\displaystyle \frac{1}{A_{m+1}^k}}=-\sum _{j=2}^n{\displaystyle \frac{A_{m+j}^k-A_{m+j-1}^k}{A_{m+j-1}^k{A}_{m+j}^k}}=-\sum _{j=2}^n{\displaystyle \frac{1}{A_{m+j-1}^k{A}_{m+j}^k}}\sum _{\nu =1}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right)A_{m+j-1}^{k-\nu }{a}_{m+j}^{\nu }.$$

This gives (4). □

Remark 1.

Setting $m=0$ and $k=1$ in (3) and (4) leads to (2).

In the next section, we apply (3) and (4) to deduce various formulas for finite sums and series which we could not locate in the literature. Our examples involve (among others) harmonic numbers, the Riemann zeta function, Fibonacci and Lucas numbers, Bernoulli polynomials, and the Pochhammer symbol. The interested reader may easily discover additional results.

2. Applications and Illustrative Examples

I. The harmonic numbers are defined by

$$H_0=0,H_n=\sum _{j=1}^n{\displaystyle \frac{1}{j}}(n\ge 1).$$

We set $a_j=1/j$. Then, we obtain from (3) with $m=0$:

$${\displaystyle \frac{1}{H_n^k}}=1+\sum _{j=2}^n{\left({\displaystyle \frac{-1}{j{H}_{j-1}{H}_j}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){(-j{H}_j)}^{\nu }.$$

If $n\to \infty$, then

$$\sum _{j=2}^{\infty }{\left({\displaystyle \frac{-1}{j{H}_{j-1}{H}_j}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){(-j{H}_j)}^{\nu }=-1.$$

(5)

From (5) with $k=2$ and $k=3$, we get

$$\sum _{j=2}^{\infty }{\displaystyle \frac{1-2j{H}_j}{{\left(j{H}_{j-1}{H}_j\right)}^2}}=-1$$

and

$$\sum _{j=2}^{\infty }{\displaystyle \frac{1-3j{H}_j+3j^2{H}_j^2}{{\left(j{H}_{j-1}{H}_j\right)}^3}}=1.$$

II. Let $s,z\in \mathbb{C}$ with $\Re \left(s\right)>1$ and $\left|z\right|\le 1$. The polylogarithm function is defined by

$${\mathrm{Li}}_s\left(z\right)=\sum _{j=1}^{\infty }{\displaystyle \frac{z^j}{j^s}}.$$

The special case $z=1$ leads to the Riemann zeta function

$${\mathrm{Li}}_s\left(1\right)=\sum _{j=1}^{\infty }{\displaystyle \frac{1}{j^s}}=\zeta \left(s\right).$$

We set

$$a_j=a_j(s,z)={\displaystyle \frac{z^j}{j^s}}\mathrm{a}\mathrm{n}\mathrm{d}{A}_j=A_j(s,z)=\sum _{\nu =1}^j{a}_{\nu }={\mathrm{Li}}_{j,s}\left(z\right).$$

Then, (3) with $m=0$ gives for $z\ne 0$,

$${\displaystyle \frac{1}{{\mathrm{Li}}_{n,s}^k\left(z\right)}}={\displaystyle \frac{1}{z^k}}+\sum _{j=2}^n{\left(-{\displaystyle \frac{z^j}{j^s{\mathrm{Li}}_{j-1,s}\left(z\right){\mathrm{Li}}_{j,s}\left(z\right)}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left(-{\displaystyle \frac{j^s}{z^j}}{\mathrm{Li}}_{j,s}\left(z\right)\right)}^{\nu }.$$

If $n\to \infty$, then we get a formula involving the polylogarithm function,

$${\displaystyle \frac{1}{{\mathrm{Li}}_s^k\left(z\right)}}={\displaystyle \frac{1}{z^k}}+\sum _{j=2}^{\infty }{\left(-{\displaystyle \frac{z^j}{j^s{\mathrm{Li}}_{j-1,s}\left(z\right){\mathrm{Li}}_{j,s}\left(z\right)}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left(-{\displaystyle \frac{j^s}{z^j}}{\mathrm{Li}}_{j,s}\left(z\right)\right)}^{\nu }.$$

(6)

From (6) with $k=1$ and $k=2$, we obtain

$${\displaystyle \frac{1}{{\mathrm{Li}}_s\left(z\right)}}={\displaystyle \frac{1}{z}}-\sum _{j=2}^{\infty }{\displaystyle \frac{z^j}{j^s{\mathrm{Li}}_{j-1,s}\left(z\right){\mathrm{Li}}_{j,s}\left(z\right)}}$$

(7)

and

$${\displaystyle \frac{1}{{\mathrm{Li}}_s^2\left(z\right)}}={\displaystyle \frac{1}{z^2}}+\sum _{j=2}^{\infty }{\displaystyle \frac{z^{2j}}{{\left(j^s{\mathrm{Li}}_{j-1,s}\left(z\right){\mathrm{Li}}_{j,s}\left(z\right)\right)}^2}}\left(1-{\displaystyle \frac{2j^s}{z^j}}{\mathrm{Li}}_{j,s}\left(z\right)\right).$$

(8)

We set $z=1$ in (7) and (8). This leads to identities involving the $\zeta$-function,

$${\displaystyle \frac{1}{\zeta \left(s\right)}}=1-\sum _{j=2}^{\infty }{\displaystyle \frac{1}{j^s{\zeta }_{j-1}\left(s\right){\zeta }_j\left(s\right)}}$$

and

$${\displaystyle \frac{1}{{\zeta }^2\left(s\right)}}=1+\sum _{j=2}^{\infty }{\displaystyle \frac{1-2j^s{\zeta }_j\left(s\right)}{{\left(j^s{\zeta }_{j-1}\left(s\right){\zeta }_j\left(s\right)\right)}^2}},$$

(9)

with ${\zeta }_j\left(s\right)={\mathrm{Li}}_{j,s}\left(1\right)={\sum }_{\nu =1}^{j}1/{\nu }^s$. In particular, from (9) with $s=2$ and $s=3$, we obtain series representations for $1/{\pi }^4$ and $1/{\zeta }^2\left(3\right)$,

$${\displaystyle \frac{36}{{\pi }^4}}=1+\sum _{j=2}^{\infty }{\displaystyle \frac{1-2j^2{H}_j^{\left(2\right)}}{{\left(j^2{H}_{j-1}^{\left(2\right)}{H}_j^{\left(2\right)}\right)}^2}}\mathrm{a}\mathrm{n}\mathrm{d}{\displaystyle \frac{1}{{\zeta }^2\left(3\right)}}=1+\sum _{j=2}^{\infty }{\displaystyle \frac{1-2j^3{H}_j^{\left(3\right)}}{{\left(j^3{H}_{j-1}^{\left(3\right)}{H}_j^{\left(3\right)}\right)}^2}},$$

where $H_n^{\left(2\right)}={\zeta }_n\left(2\right)$ and $H_n^{\left(3\right)}={\zeta }_n\left(3\right)$ are the n-th harmonic number of order 2 and 3, respectively. The number $\zeta \left(3\right)$ is known as Apéry constant.

Next, we set $z=-1$ in (7). Then,

$${\displaystyle \frac{1}{(1-2^{1-s})\zeta \left(s\right)}}=1+\sum _{j=2}^{\infty }{\displaystyle \frac{{(-1)}^j}{j^s{\mathrm{Li}}_{j-1,s}(-1){\mathrm{Li}}_{j,s}(-1)}}.$$

(10)

If $s\to 1$, then we get from (10)

$${\displaystyle \frac{1}{log2}}=1+\sum _{j=2}^{\infty }{\displaystyle \frac{{(-1)}^j}{j{S}_{j-1}{S}_j}},$$

where $S_j={\sum }_{\nu =1}^j{(-1)}^{\nu +1}/\nu$. We use (8) with $s=1$, $z=-1$. This gives

$${\displaystyle \frac{1}{{(log2)}^2}}=1+\sum _{j=2}^{\infty }{\displaystyle \frac{1+2{(-1)}^{j}j{S}_j}{{\left(j{S}_{j-1}{S}_j\right)}^2}}.$$

Finally, we apply (7) with $s=2$, $z=1/2$. We set

$${\Lambda }_j={\mathrm{Li}}_{j,2}(1/2)=\sum _{\nu =1}^j{\displaystyle \frac{1}{2^{\nu }{\nu }^2}}$$

and use

$${\mathrm{Li}}_2(1/2)={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{{log}^{2}2}{2}};$$

see Lewin [2] (p. 6). Then, we obtain

$$\sum _{j=2}^{\infty }{\displaystyle \frac{1}{2^j{j}^2{\Lambda }_{j-1}{\Lambda }_j}}=2-{\displaystyle \frac{12}{{\pi }^2-6{log}^{2}2}}.$$

III. We set

$$a_j=a_j\left(x\right)=sin\left((2j-1)x\right)\mathrm{a}\mathrm{n}\mathrm{d}{A}_{\nu }=A_{\nu }\left(x\right)=\sum _{j=1}^{\nu }{a}_j={\displaystyle \frac{{sin}^2\left(\nu x\right)}{sin\left(x\right)}};$$

see Milovanović et al. [3] (p. 326). Using (3) with $m=0$ yields

$${\displaystyle \frac{{sin}^k\left(x\right)}{{sin}^{2k}\left(nx\right)}}={\displaystyle \frac{1}{{sin}^k\left(x\right)}}+\sum _{j=2}^n{\left({\displaystyle \frac{-sin\left((2j-1)x\right){sin}^2\left(x\right)}{{sin}^2\left((j-1)x\right){sin}^2\left(jx\right)}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left({\displaystyle \frac{-{sin}^2\left(jx\right)}{sin\left(\right(2j-1\left)x\right)sin\left(x\right)}}\right)}^{\nu }.$$

We set $x=\pi /\left(3n\right)$ and $k=1$. Since $sin(\pi /3)=\sqrt{3}/2$, we obtain the summation formula

$$\sum _{j=2}^n{\displaystyle \frac{sin\left(\frac{(2j-1)\pi }{3n}\right)}{{sin}^2\left(\frac{(j-1)\pi }{3n}\right){sin}^2\left(\frac{j\pi }{3n}\right)}}={\displaystyle \frac{1}{{sin}^3\left(\frac{\pi }{3n}\right)}}-{\displaystyle \frac{4}{3sin\left(\frac{\pi }{3n}\right)}}.$$

IV. The next example offers formulas involving the binomial coefficient $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1/2}{n}}\right)$ which is connected with the classical central binomial coefficient by

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+1/2}{n}}\right)={\displaystyle \frac{2n+1}{2^{2n}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right).$$

(11)

We set

$$a_j={\displaystyle \frac{1}{2^{2j}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2j}{j}}\right)\mathrm{and}{A}_j=\sum _{\nu =0}^j{a}_{\nu }=\left({\displaystyle \genfrac{}{}{0pt}{}{j+1/2}{j}}\right);$$

see Gould [4] (p. 14). We apply (3) with $m=0$ (and with $A_j={\sum }_{\nu =0}^j{a}_{\nu }$) and use (11). Simplifying the expressions leads to

$${\displaystyle \frac{1}{{\left(\genfrac{}{}{0pt}{}{n+1/2}{n}\right)}^k}}=1-\sum _{j=1}^n{\displaystyle \frac{1}{{\left(\genfrac{}{}{0pt}{}{j-1/2}{j-1}\right)}^k}}\left\{1-{\left({\displaystyle \frac{2j}{2j+1}}\right)}^k\right\}.$$

Since

$$\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)\sim {\displaystyle \frac{4^n}{\sqrt{\pi n}}}(n\to \infty ),$$

(12)

see Olver et al. [5] (26.3.12), we conclude from (11) that

$$\underset{n\to \infty }{lim}{\displaystyle \frac{1}{\sqrt{n}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1/2}{n}}\right)={\displaystyle \frac{2}{\sqrt{\pi }}}.$$

This leads to the limit relation

$$\underset{n\to \infty }{lim}{n}^{k/2}\left(1-\sum _{j=1}^n{\displaystyle \frac{1}{{\left(\genfrac{}{}{0pt}{}{j-1/2}{j-1}\right)}^k}}\left\{1-{\left({\displaystyle \frac{2j}{2j+1}}\right)}^k\right\}\right)={\left({\displaystyle \frac{1}{2}}\sqrt{\pi }\right)}^k$$

and the series formula

$$\sum _{j=1}^{\infty }{\displaystyle \frac{1}{{\left(\genfrac{}{}{0pt}{}{j-1/2}{j-1}\right)}^k}}\left\{1-{\left({\displaystyle \frac{2j}{2j+1}}\right)}^k\right\}=1.$$

V. We set

$$a_j={\displaystyle \frac{2^{-2j}}{1-2j}}\left({\displaystyle \genfrac{}{}{0pt}{}{2j}{j}}\right)\mathrm{and}{A}_j=\sum _{\nu =1}^j{a}_{\nu }=2^{-2j}\left({\displaystyle \genfrac{}{}{0pt}{}{2j}{j}}\right)-1;$$

see Riordan [6] (p. 130). Then, we apply (3) with $m=0$, $k=1$. This gives

$$2+{\left(2^{-2n}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)-1\right)}^{-1}=\sum _{j=2}^n{\displaystyle \frac{2^{-2j}}{2j-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2j}{j}}\right){\left(2^{-2(j-1)}\left({\displaystyle \genfrac{}{}{0pt}{}{2(j-1)}{j-1}}\right)-1\right)}^{-1}{\left(2^{-2j}\left({\displaystyle \genfrac{}{}{0pt}{}{2j}{j}}\right)-1\right)}^{-1}.$$

Using

$$\left({\displaystyle \genfrac{}{}{0pt}{}{2(j+1)}{j+1}}\right)={\displaystyle \frac{2(2j+1)}{j+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2j}{j}}\right)$$

leads to

$$4+2{\left(2^{-2n}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)-1\right)}^{-1}=\sum _{j=1}^{n-1}{2}^{-2j}{C}_j{\left(2^{-2j}\left({\displaystyle \genfrac{}{}{0pt}{}{2j}{j}}\right)-1\right)}^{-1}{\left(2^{-2(j+1)}\left({\displaystyle \genfrac{}{}{0pt}{}{2(j+1)}{j+1}}\right)-1\right)}^{-1},$$

where $C_n=\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)/(n+1)$ denotes the n-th Catalan number. Next, we apply the asymptotic formula (12). This yields

$$\sum _{j=1}^{\infty }{2}^{-2j}{C}_j{\left(2^{-2j}\left({\displaystyle \genfrac{}{}{0pt}{}{2j}{j}}\right)-1\right)}^{-1}{\left(2^{-2(j+1)}\left({\displaystyle \genfrac{}{}{0pt}{}{2(j+1)}{j+1}}\right)-1\right)}^{-1}=4.$$

VI. The Pochhammer symbol (or rising factorial) is defined by

$${\left(x\right)}_n=\prod _{j=0}^{n-1}(x+j)={\displaystyle \frac{\mathrm{\Gamma }(x+n)}{\mathrm{\Gamma }\left(x\right)}}.$$

Let $p\ge 1$ be an integer. The identity

$$\sum _{\nu =1}^n{\displaystyle \frac{1}{{\left(\nu \right)}_{p+1}}}={\displaystyle \frac{1}{p·p!}}\left(1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{n+p}{p}\right)}}\right)$$

is given in Cheon and El-Mikkawy [7]. We set

$$a_j=a_j\left(p\right)={\displaystyle \frac{1}{{\left(j\right)}_{p+1}}}.$$

Then, we obtain

$$A_j=A_j\left(p\right)=\sum _{\nu =1}^j{a}_{\nu }={\displaystyle \frac{1}{p·p!}}\left(1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+p}{p}\right)}}\right).$$

Applying (4) with $m=0$ gives

$${\left\{{\displaystyle \frac{1}{p·p!}}\left(1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{n+p}{p}\right)}}\right)\right\}}^{-k}={\left((p+1)!\right)}^k-\sum _{j=2}^n{\left({\displaystyle \frac{p·p!}{1-\frac{1}{\left(\genfrac{}{}{0pt}{}{j+p}{p}\right)}}}\right)}^k\sum _{\nu =1}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left({\displaystyle \frac{1}{{\left(j\right)}_{p+1}}}{\displaystyle \frac{p·p!}{1-\frac{1}{\left(\genfrac{}{}{0pt}{}{j-1+p}{p}\right)}}}\right)}^{\nu }.$$

For $k=1$, we obtain

$${\left(1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{n+p}{p}\right)}}\right)}^{-1}={\displaystyle \frac{p+1}{p}}-p·p!\sum _{j=2}^n{\displaystyle \frac{1}{{\left(j\right)}_{p+1}}}{\left(1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+p}{p}\right)}}\right)}^{-1}{\left(1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j-1+p}{p}\right)}}\right)}^{-1}.$$

We let $n\to \infty$ and simplify. This leads to

$${\displaystyle \frac{1}{p^2·p!}}=\sum _{j=2}^{\infty }{\displaystyle \frac{1}{{\left(j\right)}_{p+1}}}{\left(1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j+p}{p}\right)}}\right)}^{-1}{\left(1-{\displaystyle \frac{1}{\left(\genfrac{}{}{0pt}{}{j-1+p}{p}\right)}}\right)}^{-1}.$$

VII. The Bernoulli polynomials $B_n\left(x\right)$ are defined by the generating function

$${\displaystyle \frac{t{e}^{tx}}{e^t-1}}=\sum _{n=0}^{\infty }{B}_n\left(x\right){\displaystyle \frac{t^n}{n!}}.$$

The numbers $B_n=B_n\left(0\right)$ are the well-known Bernoulli numbers.

(i) Let $p\ge 2$ be an integer. We set

$$a_j=a_j\left(p\right)=j^{p-1}\mathrm{and}{A}_j=A_j\left(p\right)=\sum _{\nu =1}^j{a}_{\nu }={\displaystyle \frac{B_p(j+1)-B_p}{p}};$$

see Abramowitz and Stegun [8] (p. 804). Applying (4) with $m=0$ gives

$${\left({\displaystyle \frac{p}{B_p(n+1)-B_p}}\right)}^k=1-\sum _{j=2}^n{\left({\displaystyle \frac{p}{B_p(j+1)-B_p}}\right)}^k\sum _{\nu =1}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left({\displaystyle \frac{j^{p-1}p}{B_p\left(j\right)-B_p}}\right)}^{\nu }.$$

(13)

For $k=1$, we get

$${\displaystyle \frac{p}{B_p(n+1)-B_p}}=1-p^2\sum _{j=2}^n{\displaystyle \frac{j^{p-1}}{\left(B_p(j+1)-B_p\right)\left(B_p\left(j\right)-B_p\right)}}.$$

We let $n\to \infty$ and use ${lim}_{x\to \infty }{B}_p\left(x\right)=\infty$. Then, we obtain

$${\displaystyle \frac{1}{p^2}}=\sum _{j=2}^{\infty }{\displaystyle \frac{j^{p-1}}{\left(B_p(j+1)-B_p\right)\left(B_p\left(j\right)-B_p\right)}}.$$

Let $p\ge 3$ be odd. Then, $B_p=0$. We set $k=2$ and let $n\to \infty$. Then, we conclude from (13) that

$${\displaystyle \frac{1}{p^3}}=2\sum _{j=2}^{\infty }{\displaystyle \frac{j^{p-1}}{B_p\left(j\right)B_p{(j+1)}^2}}+p\sum _{j=2}^{\infty }{\left({\displaystyle \frac{j^{p-1}}{B_p\left(j\right)B_p(j+1)}}\right)}^2.$$

(ii) Let $p\ge 0$ be even. We set

$$a_j=a_j\left(p\right)={(2j-1)}^p\mathrm{and}{A}_j=A_j\left(p\right)=\sum _{\nu =1}^j{a}_{\nu }={\displaystyle \frac{2^p}{p+1}}{B}_{p+1}\left(j+{\displaystyle \frac{1}{2}}\right),$$

see Kilar et al. [9], and apply (4) with $m=0$. This gives

$${\left({\displaystyle \frac{p+1}{2^p{B}_{p+1}(n+1/2)}}\right)}^k=1-\sum _{j=2}^n{\left({\displaystyle \frac{p+1}{2^p{B}_{p+1}(j+1/2)}}\right)}^k\sum _{\nu =1}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left({\displaystyle \frac{{(2j-1)}^p(p+1)}{2^p{B}_{p+1}(j-1/2)}}\right)}^{\nu }.$$

For $k=1$, we get

$${\displaystyle \frac{p+1}{2^p{B}_{p+1}(n+1/2)}}=1-{\displaystyle \frac{{(p+1)}^2}{4^p}}\sum _{j=2}^n{\displaystyle \frac{{(2j-1)}^p}{B_{p+1}(j+1/2)B_{p+1}(j-1/2)}}.$$

If $n\to \infty$, then

$$\sum _{j=2}^{\infty }{\displaystyle \frac{{(2j-1)}^p}{B_{p+1}(j+1/2)B_{p+1}(j-1/2)}}={\displaystyle \frac{4^p}{{(p+1)}^2}}.$$

VIII. The classical Fibonacci numbers are defined by the recurrence relation

$$F_0=0,F_1=1,F_n=F_{n-1}+F_{n-2}(n\ge 2),$$

whereas the Lucas numbers are given by

$$L_0=2,L_1=1,L_n=L_{n-1}+L_{n-2}(n\ge 2).$$

The numbers $F_n$ and $L_n$ are related by numerous identities, like, for example,

$$F_{2n}=F_n{L}_n,L_n=F_{n-1}+F_{n+1}=2F_{n+1}-F_n.$$

(i) We set

$$a_j=F_j\mathrm{and}{A}_j=\sum _{\nu =1}^j{F}_{\nu }=F_{j+2}-1;$$

see Rosen at al. [10] (Section 3.1.2). Applying (3) with $m=0$ gives

$${\displaystyle \frac{1}{{(F_{n+2}-1)}^k}}=1+\sum _{j=2}^n{\left({\displaystyle \frac{-F_j}{(F_{j+1}-1)(F_{j+2}-1)}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left({\displaystyle \frac{1-F_{j+2}}{F_j}}\right)}^{\nu }.$$

(14)

In particular, for $k=1$, $k=2$, and $k=3$, we obtain the formulas

$${\displaystyle \frac{1}{F_{n+2}-1}}=1-\sum _{j=2}^n{\displaystyle \frac{F_j}{(F_{j+1}-1)(F_{j+2}-1)}},$$

$${\displaystyle \frac{1}{{(F_{n+2}-1)}^2}}=1-\sum _{j=2}^n{\displaystyle \frac{F_j(F_{j+3}-2)}{{(F_{j+1}-1)}^2{(F_{j+2}-1)}^2}},$$

$${\displaystyle \frac{1}{{(F_{n+2}-1)}^3}}=1-\sum _{j=2}^n{\displaystyle \frac{F_j\left(F_j^2+3(F_{j+1}-1)(F_{j+2}-1)\right)}{{(F_{j+1}-1)}^3{(F_{j+2}-1)}^3}}.$$

We let $n\to \infty$. Then, (14) leads to

$$\sum _{j=2}^{\infty }{\left({\displaystyle \frac{-F_j}{(F_{j+1}-1)(F_{j+2}-1)}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left({\displaystyle \frac{1-F_{j+2}}{F_j}}\right)}^{\nu }=-1.$$

For $k=1$, $k=2$, and $k=3$, we obtain

$$\sum _{j=2}^{\infty }{\displaystyle \frac{F_j}{(F_{j+1}-1)(F_{j+2}-1)}}=1,$$

$$\sum _{j=2}^{\infty }{\displaystyle \frac{F_j(F_{j+3}-2)}{{(F_{j+1}-1)}^2{(F_{j+2}-1)}^2}}=1,$$

$$\sum _{j=2}^{\infty }{\displaystyle \frac{F_j\left(F_j^2+3(F_{j+1}-1)(F_{j+2}-1)\right)}{{(F_{j+1}-1)}^3{(F_{j+2}-1)}^3}}=1.$$

(ii) We set

$$a_j=F_j^2\mathrm{and}{A}_j=\sum _{\nu =1}^j{a}_{\nu }=F_j{F}_{j+1};$$

see Rosen et al. [10] (Section 3.1.2). We apply (4) with $m=0$. Then,

$${\displaystyle \frac{1}{{\left(F_n{F}_{n+1}\right)}^k}}=1-\sum _{j=2}^n{\displaystyle \frac{1}{{\left(F_j{F}_{j+1}\right)}^k}}\sum _{\nu =1}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left({\displaystyle \frac{F_j}{F_{j-1}}}\right)}^{\nu }.$$

For $k=2$ and $k=3$, we obtain

$${\displaystyle \frac{1}{F_n^2{F}_{n+1}^2}}=1-\sum _{j=2}^n{\displaystyle \frac{F_{j-1}+F_{j+1}}{F_{j-1}^2{F}_j{F}_{j+1}^2}},$$

$${\displaystyle \frac{1}{F_n^3{F}_{n+1}^3}}=1-\sum _{j=2}^n{\displaystyle \frac{3F_{j-1}{F}_{j+1}+F_j^2}{F_{j-1}^3{F}_j^2{F}_{j+1}^3}}.$$

And, if $n\to \infty$, then

$$\sum _{j=2}^{\infty }{\displaystyle \frac{1}{F_{j-1}{F}_j{F}_{j+1}^2}}+\sum _{j=2}^{\infty }{\displaystyle \frac{1}{F_{j-1}^2{F}_j{F}_{j+1}}}=1,$$

$$3\sum _{j=2}^{\infty }{\displaystyle \frac{1}{{\left(F_{j-1}{F}_j{F}_{j+1}\right)}^2}}+\sum _{j=2}^{\infty }{\displaystyle \frac{1}{{\left(F_{j-1}{F}_{j+1}\right)}^3}}=1.$$

(iii) We set

$$a_j={(-1)}^j{F}_{2j}\mathrm{and}{A}_j=\sum _{\nu =1}^j{a}_{\nu }={(-1)}^j{F}_j{F}_{j+1};$$

see Adegoke et al. [11] and Kılıç et al. [12]. Applying (3) with $m=0$ gives

$${\displaystyle \frac{{(-1)}^{nk}}{{\left(F_n{F}_{n+1}\right)}^k}}={(-1)}^k+\sum _{j=2}^n{\left({(-1)}^j{\displaystyle \frac{L_j}{F_{j-1}{F}_j{F}_{j+1}}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){\left(-{\displaystyle \frac{F_{j+1}}{L_j}}\right)}^{\nu }.$$

(15)

For $k=1$, we obtain

$${\displaystyle \frac{{(-1)}^n}{F_n{F}_{n+1}}}=-1+\sum _{j=2}^n{(-1)}^j{\displaystyle \frac{L_j}{F_{j-1}{F}_j{F}_{j+1}}}$$

(16)

and using

$$L_j^2-2L_j{F}_{j+1}=-F_{2j}$$

we get from (15) with $k=2$

$${\displaystyle \frac{1}{F_n^2{F}_{n+1}^2}}=1-\sum _{j=2}^n{\displaystyle \frac{F_{2j}}{F_{j-1}^2{F}_j^2{F}_{j+1}^2}}.$$

(17)

If $n\to \infty$, then, (16) and (17) lead to

$$\sum _{j=2}^{\infty }{(-1)}^j{\displaystyle \frac{L_j}{F_{j-1}{F}_j{F}_{j+1}}}=1$$

and

$$\sum _{j=2}^{\infty }{\displaystyle \frac{F_{2j}}{F_{j-1}^2{F}_j^2{F}_{j+1}^2}}=1.$$

(iv) We set $a_j=1/F_{2^j}$. Since

$$\sum _{\nu =0}^j{a}_{\nu }=3-{\displaystyle \frac{F_{2^j-1}}{F_{2^j}}}(j\ge 1),$$

see Rabinowitz [13], we obtain

$$A_j=\sum _{\nu =1}^j{a}_{\nu }={\displaystyle \frac{L_{2^j-1}}{F_{2^j}}}.$$

We apply (3) with $m=0$. Then,

$${\left({\displaystyle \frac{F_{2^n}}{L_{2^n-1}}}\right)}^k=1+\sum _{j=2}^n{\left(-{\displaystyle \frac{F_{2^{j-1}}}{L_{2^{j-1}-1}{L}_{2^j-1}}}\right)}^k\sum _{\nu =0}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\nu }}\right){(-L_{2^j-1})}^{\nu }.$$

(18)

We set $k=1$. This yields

$${\displaystyle \frac{F_{2^n}}{L_{2^n-1}}}={\displaystyle \frac{3}{2}}-\sum _{j=0}^{n-1}{\displaystyle \frac{F_{2^j}}{L_{2^j-1}{L}_{2^{j+1}-1}}}.$$

We let $n\to \infty$ and apply the limit relation

$$\underset{n\to \infty }{lim}{\displaystyle \frac{F_n}{L_{n-1}}}=\underset{n\to \infty }{lim}{\displaystyle \frac{1}{5}}\left({\displaystyle \frac{L_{n+1}}{L_{n-1}}}+1\right)={\displaystyle \frac{\varphi }{\sqrt{5}}},$$

where $\varphi =(1+\sqrt{5})/2$ is the golden ratio. Then,

$${\displaystyle \frac{1}{2\sqrt{5}}}=1-\sum _{j=0}^{\infty }{\displaystyle \frac{F_{2^j}}{L_{2^j-1}{L}_{2^{j+1}-1}}}.$$

If we use (18) with $k=2$, then we obtain the formulas

$${\displaystyle \frac{F_{2^n}^2}{L_{2^n-1}^2}}={\displaystyle \frac{5}{4}}+\sum _{j=0}^{n-1}{\displaystyle \frac{F_{2^j}^2(1-2L_{2^{j+1}-1})}{{\left(L_{2^j-1}{L}_{2^{j+1}-1}\right)}^2}}$$

and

$${\displaystyle \frac{1}{2\sqrt{5}}}={\displaystyle \frac{19}{20}}+\sum _{j=0}^{\infty }{\displaystyle \frac{F_{2^j}^2(1-2L_{2^{j+1}-1})}{{\left(L_{2^j-1}{L}_{2^{j+1}-1}\right)}^2}}.$$