\section{Introduction and preliminaries }


The theory of transcendental numbers has a long history. We know since J. Liouville in 1844 that the very rapidly converging sequences of rational numbers provide examples of transcendental numbers. So, in his famous paper [6], Liouville showed that a real number admitting very good rational approximation can not be algebraic, then he explicitly constructed the first examples of transcendental numbers.\\

There are a number of sufficient conditions known within the literature for an infinite series, $ \sum_{n = 1}^{\infty} 1/a_n$, of positive rational numbers to converge to an irrational number, see [2, 8, 10].These conditions, which are quite varied in form, share one common feature, namely, they all require rapid growth of the sequence ${ a_n }$ to deduce irrationality of the series. As an illustration consider the following results of J. Sandor which have been taken from [10] and [11].\\

From this direction, the transcendence of some infinite series has been studied by several authors such as M.A. Nyblom [7], J. Hančl and J. Štěpnička [4]. we also note that the transcendence of some power series with rational coefficients is given by some authors such as J. P. Allouche [1] and G. K. Gözer [3].



\newpage
\begin{ThR}
\normalfont
 Let $ \alpha $ be a real number, $ \delta $ a real number > 2, if there exists an infinity rational numbers $ \dfrac{p}{q} $ with $ gcd(p, q) = 1 $ such that $$ \left| \alpha - \frac{p}{q} \right| < \frac{1}{q^{ \delta} }, $$

then $ \alpha $ is a transcendental number.
\end{ThR}



\poubelle{
\begin{Th}\textbf{( Eisenstein, 1852), ( Heine, 1853 )  }~\\

Any algebraic power series $ f = \sum_{n\geq 0} a_nt^n $ in $ \mathbb{Q}\left[\left[t\right]\right]$ is globally bounded :  There exists an integer $ C > 0 $ such that $ a_nC^n $ is an integer for all $ n \geq 1 $.

\end{Th}

Consequently, the power series:

\begin{enumerate}[label= \textbf{\roman*)}]
    \item $ \sum_n \frac{t^n}{n} $
    \item $ \sum_n \frac{t^n}{n^2 + 1 }$
    \item $ \sum_n \frac{t^n}{n!} $
\end{enumerate}
 are transcendental.
\begin{Rems}~\\
\begin{enumerate}[label= \textbf{\bullet}]
    \item The smallest possible constant C is called Eisenstein constant of $ f $.
    \item Best current bound is, see $\left[ Dwork,\hspace{0.1cm} van \hspace{0.1cm} der \hspace{0.1cm} Poorten \hspace{0.1cm} 1992 \right]$ $$ C \leq 4.8 \left(8e^{-3}D^{4 + 2.74 \log D} e^{1.22D} \right)^D. H^{2D - 1} = e^{O\left(D^2\right)}. H^{2D - 1} $$
    where $ D $ is the degree of the (minpoly of) $ f ,$ and $ H $ the max of its coefficients.
\end{enumerate}


\end{Rems}
}
\poubelle{
\begin{Rem}
The infinite series :

\begin{enumerate}[label= \textbf{\roman*)}]
\item $ \sum_{n \geq 0} \frac{1}{n!} = e,$
\item  $ \sum_{n \geq 0} \frac{1}{n^2 + 1} = \frac{\pi^2}{2} $

\end{enumerate}
are transcendental numbers.

\end{Rem}
}

\section{Main results}

\subsection{Transcendence}
The first main result is given in the following theorem.

\begin{Th}{\it  Let $(a_n)_{n \geq 1}$ be a sequence of non-zero natural integers and $ \alpha $ be a positive real\\ > 2 such that
\begin{equation}
a_{n + 1} > a_n^{\alpha + 1}, \hspace{0.1cm}\text{ for all } n \geq 1. \\
\end{equation}
 Then the series $ \sum_{n = 1}^ \infty 1/{a_n} $ converges to a transcendental number.}
\end{Th}

 In order to prove Theorem 1, we need some preliminary lemmas.
\begin{Lem}{\it
Let $$ \frac{p_m}{q_m} = \sum_{k = 1}^m \frac{1}{a_k} $$ such that $ (p_m, q_m) = 1$. Then, we have
\begin{equation}
   q_m \leq a_1a_2...a_m.
\end{equation}}
\end{Lem}

\begin{proof}
Since $ (p_m, q_m) = 1$,  the lowest common denominator of the fraction $\dfrac{1}{a_1}...\dfrac{1}{a_m}$ must be greater than or equal to $ q_m $. So we deduce $ q_m \leq a_1a_2...a_m$.

\end{proof}

\begin{Lem}{\it
Let $ (a_n)_{n \geq 1}$ be a sequence of natural integers > 0, and $\alpha$ be a given real > 2. The hypothesis $ a_{n+1} > a_n^{\alpha + 1}$ implies that
\begin{equation}
    \lim\limits_{n \rightarrow \infty} \frac{\left(a_1a_2...a_n \right)^\alpha}{a_{n+1}} = 0
\end{equation}}

\end{Lem}

\begin{proof}
We put $$ b_n = \frac{\left(a_1a_2...a_n \right)^{\alpha}}{a_{n+1}}, $$ and we show that $ \lim\limits_{n \rightarrow \infty} b_n = 0.$ We have
\begin{align*}
\ln \left(\frac{1}{b_n} \right) &= \ln \left(a_{n+1} \right) - \alpha \sum_{k = 1}^n \ln \left(a_k\right)\\
&= \sum_{k=1}^n \ln \left(\frac{a_{k+1}}{a_k} \right) + \ln\left(a_1\right) - \alpha \sum_{k=1}^n \ln\left(a_k\right)\\
&= \sum_{k=1}^n \ln\left(\frac{a_{k+1}}{a_k^{\alpha + 1}} \right) + \ln \left(a_1\right)\\
&\geq \sum_{k=1}^n \ln\left(\frac{a_{k+1}}{a_k^{\alpha + 1}}\right).
\end{align*}
Since $ \dfrac{a_{n+1}}{a_{n}^{\alpha +1}} > 1, $ then there exists $  \delta > 0 $ such that $ \dfrac{a_{n+1}}{a_{n}^{\alpha +1}} > 1 + \delta.$ Therefore, we get
\begin{equation*}
\ln\left(\frac{1}{b_n}\right) \geq n \ln\left( 1 + \delta \right).
\end{equation*}

From this, we deduce that, $ \lim\limits_{n \rightarrow +\infty} \ln\left(\dfrac{1}{b_n}\right) = +\infty  $, then $ \lim\limits_{n \rightarrow +\infty} b_n = 0$.\\
\end{proof}

\begin{p}\normalfont
According to the hypothesis, the series $ \sum_n \dfrac{1}{a_n}$ is convergent.\\
Put $ \theta = \sum_{n = 1} ^\infty \dfrac{1}{a_n} $ and $ \dfrac{p_m}{q_m} = \sum_{n = 1}^m \dfrac{1}{a_n} $. From the equality,

\begin{equation*}
\left| \theta - \frac{p_m}{q_m} \right| = \sum_{n = m +1}^\infty \frac{1}{a_n},
\end{equation*}
we obtain
\begin{equation*}
q_m^\alpha \left| \theta - \frac{p_m}{q_m} \right| = \sum_{n = m + 1}^\infty \frac{q_m^\alpha}{a_n}. \\
\end{equation*}
The relationship (2) implies that
\begin{equation*}
q_m^\alpha \left| \theta - \frac{p_m}{q_m} \right| \leq \left( a_1a_2...a_m \right)^\alpha
\end{equation*}
\begin{equation*}
\sum_{n = m + 1}^\infty \frac{1}{a_n} \leq b_m \sum_{n = m + 1}^\infty \frac{a_{m+1}}{a_n},
\end{equation*}
with $ b_m = \dfrac{\left(a_1a_2...a_m \right)^\alpha}{a_{m+1}}.$\\
Furthermore, we have
\begin{equation}
    \frac{a_n}{a_{n+1}} < \frac{1}{a_n^\alpha} < \frac{1}{a_n}, \hspace{0.08cm} all\hspace{0.08cm} n \geq 1.
\end{equation}
Then, we obtain
\begin{align*}
    q^\alpha_m \left| \theta - \frac{p_m}{q_m} \right| &< b_m \left( 1 + \sum_{k = 1}^\infty \frac{a_{m+1}}{a_{m+k+1}} \right)\\
    &< b_m \left( 1 + \sum_{k=1}^\infty \frac{a_{m+k}}{a_{m+k+1}} \right)\\
    &< b_m \left( 1 + \sum_{k=1}^\infty \frac{1}{a_{m+k}} \right) \\
    &< b_m \left( 1 + \theta \right). \hspace{2cm}
\end{align*}

According to the relationship (3), and for m sufficiently large, we get $ b_m < \left(1 + \theta \right)^{-1}.$\\

Therefore for m sufficiently large, we have $$ q_m^\alpha \left| \theta - \frac{p_m}{q_m}\right| < 1. $$

Finally we find
\begin{equation}
   \left| \theta - \frac{p_m}{q_m}\right| < \frac{1}{q_m^\alpha}.
\end{equation}


Since $ \alpha > 2 $, then by Roth's theorem, $ \theta $ is a transcendental number.\\

\end{p}
\poubelle{
\chapter*{Application}
\addcontentsline{toc}{chapter}{Application}


we shall demonstrate via theorem 1, the transcendence of a family of series involving the generalised Fibonacci and Lucas sequences, denoted by $ U_n $ and $ V_n $ respectively.  Let $ \left(P, Q\right)$ be a relatively prime pair of integers such that the roots $ \alpha $ and $ \beta $ of $ x^2 - Px + Q = 0 $ are distinct, then $U_n, V_n$ are given by $$ U_n = \frac{\alpha^n - \beta^n}{\alpha - \beta} \hspace{1cm} and \hspace{1cm}  V_n = \alpha^n + \beta^n $$

It is well known that when the discriminant $ \Delta = P^2 - 4Q > 0$  both $\{U_n\}$ and $ \{V_n\} $ are increasing sequences of positive integers. In particular, for $ \left(P,Q\right) = \left(1,-1\right)$ one has $ U_n = F_n $ and $ V_n = L_n,$ where $F_n$ and $L_n$ are the Fibonacci and Lucas numbers respectively. The irrationality of the following series $$ \sum_{n = 1}^\infty \frac{1}{U_{f(n)}},$$

where $ f: \mathbb{N} \longrightarrow \mathbb{N}$ satisfies $ f( n + 1 ) \geq 2f(n),$ was proved by McDaniel in ... This result however has been proved more generally by Badea i ... for sequence $ \{x_n\}$ generated from a second order recurrence relation of the form $x_{n+1} = ax_n + bx_{n-1}$, where $x_0 = 0, x_1 = 1 $ and a, b are fixed positive integers. In particular, it was shown that if $n(k)$ is a positive integer sequence such that $ n(k+1) \geq 2n(k)-1$ for k sufficiently large then $\sum_{k=1}^\infty \frac{1}{x_{n(k)}}$ is irrational. A similar result has been found in ... were irrationality of the series $ \sum_{n=1}^\infty 1/H\left( f(n) \right) $ was established for $ \{H(k)\}$ a sequence of positive integers satisfying a homogeneous linear recurrence relation and $f(.)$ an integer valued function such that $ f(n+1) - 2f(n) \longrightarrow \infty$ as $ n \longrightarrow \infty$ and $f(n+1) \geq f(n)^2$ for all sufficiently large n. It should further be noted, that analogous irrationality results have also been found in ..., for series whose terms are reciprocals of elements of an integer sequence generated from a linear recurrence of order $ k \geq 2.$ However, the sufficiency conditions obtained there were of a more complex nature than those described in above. To contrast these general results of ..., we now establish the transcendence of $\sum_{n=1}^\infty 1/U_{f(n)}$ and $\sum_{n=1}^\infty 1/V_{f(n)},$ where the index function $f(.)$ satisfies for a fixed $ \lambda > 2$ the inequality $ f(n+1) \geq (\lambda + 2 ) f(n).$

\begin{Cor}
Let $\left(P,Q\right)$ be a relatively prime pair of integers with $ P > \left| Q + 1 \right| $ and $ Q \neq 1$ and $ \{U_n\}, \{V_n\} $ the associated generalised Fibonacci and Lucas sequences. If, for a fixed\\
\lambda > 2, the function $ f: \mathbb{N} \Longrightarrow \mathbb{N}$ has the property that $f(n+1) \geq \left(\lambda + 2\right) f(n) $, then the series $ \sum_{n=1}^\infty 1/a_n$ converges to a transcendental number, where $ a_n = U_{f(n)} $ or  $ a_n = V_{f(n)}. $
\end{Cor}

\begin{Rem}
We note that the restriction on Q is required as the sequence $\{U_n\}$ will contain infinitely many zero elements when $\left(P,Q\right) = \left(1,-1\right). $
\end{Rem}

\begin{proof}
In view of Theorem... it will suffice to demonstrate in either case that\\
$ a_{n+1}/a_n^{\lambda +1} \longrightarrow \infty$ as $n \longrightarrow \infty $. Clearly, from definition $\alpha = \left( P + \sqrt{\Delta} \right)/2$ and $ \beta =  \left( P - \sqrt{\Delta} \right)/2 $ where $ \Delta = P^2 - 4Q.$ Now from assumption $ \sqrt{\Delta} > \sqrt{\left( Q + 1 \right)^2 - 4Q } = \left| Q - 1 \right| > 1 $ and so $$ \left|\beta\right| = \left| \frac{P - \sqrt{\Delta}}{2} \right| = \frac{\left|2Q\right|}{P + \sqrt{\Delta}} < \frac{\left|2Q\right|}{\left| Q + 1 \right| + \left| Q - 1 \right|} = 1.$$
noting here that the right hand equality holds for all $Q \in \mathbb{R} $ with $\left| Q \right| \geq 1.$ Consequently, $ \left| \alpha \right| = \left|Q \right|/\left|\beta\right| > \left| Q \right| \geq 1$ and $ \left| \beta / \alpha \right| < 1.$ Now, in the case when $ a_n = U_{f(n)}$ observe $$ \frac{a_{n+1}}{a_n^{\lambda + 1}} = \alpha^{f(n+1) - (\lambda + 1 ) f(n)}\frac{\left(1 - \left(\beta / \alpha \right)^{f(n+1)} \right)}{\left(1 - \left(\beta / \alpha \right)^{f(n)} \right)^{\lambda + 1 }} \left(\sqrt{\Delta} \right)^\lambda \sim \alpha^{f(n+1) - (\lambda + 1 ) f(n)} \left(\sqrt{\Delta} \right)^\lambda, $$ as $ n \longrightarrow \infty $. While in the latter case $$ \frac{a_{n+1}}{a_n^{\lambda + 1}} = \alpha^{f(n+1) - (\lambda + 1 ) f(n)}\frac{\left(1 + \left(\beta / \alpha \right)^{f(n+1)} \right)}{\left(1 + \left(\beta / \alpha \right)^{f(n)} \right)^{\lambda + 1 }} \sim \alpha^{f(n+1) - (\lambda + 1 ) f(n)}, $$ as $ n \longrightarrow \infty. $ However as $ f(n+1) - (\lambda + 1) f(n) \geq f(n) $ and $ \alpha > 1 $ one has $ \alpha^{f(n)} \longrightarrow \infty $ as $ n \longrightarrow \infty ,$ hence the condition of Theorem ... is satisfied.
\end{proof}

In ... Badea showed that $\sum_{n=0}^\infty 1/F_{2^n} = 7 - \sqrt{5}/2 $ and so the series sums to an irrational (but algebraic) number. By using Corollary ... one can now demonstrate the transcendence of a similar series formed by replacing the index function $ 2^n$ with $5^n$. Indeed when $ \left(P,Q\right) = \left(1,-1\right) $ one obtains for $ \lambda = 3 $ and $f(n+1) = 5f(n), $ with $f(1) = 5,$ the transcendence of both $\sum_{n=1}^\infty 1/F_{5^n} $ and $ \sum_{n=1}^\infty 1/L_{5^n}. $ Badea also established the irrationality of both $\sum_{n=1}^\infty 1/F_{2^n + 1} $ in ... and $ \sum_{n=1}^\infty 1/F_{t2^n} $  (for $t \in \mathbb{N} $ fixed ) in ..., although the latter result can be found in ... and in ... for the case of a Lucas sequence. In connection with these results we can again use the previous corollary with $ f(n) = 5^n + 1 $ and $ f(n) = t5^n $ together with $\lambda = 2.5 $ to now deduce transcendence of the series  $ \sum_{n=1}^\infty 1/F_{5^n +1} $ and  $ \sum_{n=1}^\infty 1/tF_{5^n} $ together with $ \sum_{n=1}^\imfty 1/L_{5^n}$ and $ \sum_{n=1}^\imfty 1/L_{t5^n}.$ \\
To conclude we shall examine yet another application of theorem ... in generating alternate families of transcendental valued series. Mahler and Mignotte in .. and .. independently proved the transcendental of the series $ \sum_{n=1}^\infty 1/n!F_{2^n}. $ Unfortunately this result cannot be obtained via theorem .. since for any $\lambda > 2 $  $$ \liminf_{n \longrighttarrow \infty} \frac{\left( n + 1 \right)! F_{2^{n+1}}}{ \left( n! F_{2^n}\right)^{\left(\lambda + 1\right)}} = 0 $$

However, if we interchange the role of $ 2^n$ and $ n! $ in the previous series one can now prove via Theorem ... the transcendence of $ \sum_{n = 1}^\infty 1/2^nF_{n!}$. We shall demonstrate this by first establishing the following general result.

\begin{Cor}
Let $\left(P,Q\right) $ be a relatively prime pair of integers with $ P > \left| Q + 1 \right| $ and $ Q \neq 1 $ and $ \{ U_n \} $ and $ \{ V_n \} $ the associated generalised Fibonacci and Lucas sequences. if in addition m is any positive integer greater than unity then the series $\sum_{n=1}^\infty 1/a_n $ converges to a transcendental number, where $ a_n = m^nU__{n!} $ or $ a_n = m^nV_{n!}.$
\end{Cor}

\begin{proo}

\end{proo}
}

We will now give a corollary as an application of our main result.
\begin{Cor}{\it
Any subseries of the series $\sum_{n \geq 1} \frac{1}{a_n}$, where the terms $ a_n \in \mathbb{N}\setminus \{0\} $ satisfy (1) will have a transcendental sum.}
\end{Cor}

\begin{proof}
Consider an arbitrary subseries $ \sum_{n \geq 1 } 1/c_n $ then by definition there must exist a strictly monotone increasing function $ g : \mathbb{N} \rightarrow \mathbb{N}$ such that $ c_n = a_{g(n)} .$ Clearly as $ g\left( n + 1 \right) \geq g(n) + 1 $ one has $$ \frac{c_{n+1}}{c_n^{\alpha + 1}} = \frac{a_{g(n+1)}}{a_{g(n)}^{\alpha + 1 }} \geq \frac{a_{g(n) + 1 }}{a_{g(n)}^{\alpha + 1}} > 1, $$
 and  by Theorem 1 the subseries has a transcendental sum.
\end{proof}

\begin{Exp}\normalfont
We consider the following sequence: $$ \left\{
\begin{array}{rcr}
a_n = 2^{n! + 1}, & n \geq 1 \\
a_{n+1} > a_n^{3+\epsilon}, & n \geq 3
\end{array}
\right. $$

By applying Theorem 1, the series $$ \sum_n \frac{1}{2^{n! + 1}} $$  converge to a transcendental number.


\end{Exp}


\subsection{Transcendental measure}
The second main result of this paper is to give a transcendental measure of $ \theta = \sum_{n=1}^{\infty} \frac{1}{a_n} $. In this subsection, we keep the same notations as in the first subsection.\\

\begin{Th}
Let $P \in \mathbb{Z}[X]\textbackslash \left\{  0 \right \} $ be a polynomial of degree $ d \geq 2$ and height $ H $. Let $\alpha > d $ and $k > 1$ be two real numbers such that  $$ a_n^{\alpha + 1 } \leq a_{ n + 1 } < a_n^{k \alpha }, \hspace{0.08cm}\text{ for all } n \geq 1. $$
Then, we have $$ \lvert P(\theta ) \rvert > \dfrac{1}{\left(Hd\left(d+1\right)\right)^{\frac{kd\left(\alpha + 1\right)}{\alpha - d}}}. $$
\end{Th}

 To prove this Theorem, we need the following lemma.\\

\begin{Lem}
\begin{enumerate}[label = \textbf{(\roman*)}] \item The hypothesis  $\hspace{0.2cm} a_n^{\alpha + 1 } \leq  a_{ n + 1 }$ implies that
    \begin{equation}
        q_n \leq a_n^{\frac{\alpha + 1 }{\alpha}}, for\hspace{0,1cm}  n \geq 1.
    \end{equation}
    \item The hypothesis $\hspace{0.2cm} a_{ n + 1 } <a_n^{k \alpha } $ implies that
    \begin{equation}
        q_{n+1} < q_n^{k(\alpha + 1 )}, for  \hspace{0,1cm} n \geq 1.
    \end{equation}
\end{enumerate}
\end{Lem}

\begin{proof}
\begin{enumerate}[label = \textbf{\roman*)}]\item The hypothesis of i) implies that  $$ \hspace{-1.7cm}a_n \leq a_{n + 1 }^{\frac{1}{\alpha + 1}} . $$ Then for all $ 1 \leq j \leq n-1,$ we obtain $$ a_j \leq a_n^{\left(\frac{1}{\alpha +1 }\right)^
{n-j}}. $$
On the other hand, according to the relationship (2), one has $$\hspace{-3cm} q_n \leq a_1...a_{n-1}.a_n, $$ this implies
    $$ \hspace{-2cm} q_n \leq a_n^{1 + \frac{1}{\alpha + 1} + \frac{1}{\left( \alpha + 1 \right)^2} + ... + \frac{1}{\left( \alpha + 1 \right)^{n - 1}}}.$$ Which gives
    $$  \hspace{-2cm} q_n \leq a_n^{\frac{1}{1 - \frac{1}{\alpha + 1}}}. $$
    Finally we obtain $$ \hspace{0.8cm} q_n \leq a_n^{\frac{\alpha + 1 }{\alpha}}, \hspace{0.5cm} for \hspace{0.1cm} all \hspace{0.1cm} n \geq 1. $$
\item According to the relationship (6), we have
\begin{align*}
    q_n &\leq a_n^{\frac{\alpha + 1 }{\alpha}} < a_{n-1}^{\frac{\alpha +1}{\alpha}k\alpha} = a_{n-1}^{k\left( \alpha + 1 \right)}.
\end{align*}
Since $a_n < q_n$ for all $ n \geq 1$, we obtain
$$ q_n < q_{n - 1}^{k\left(\alpha + 1 \right)}. $$
\end{enumerate}

\end{proof}

\begin{pf}\normalfont
Put $$  \theta_n = \frac{p_n}{q_n} = \sum_{k=1}^{n} \frac{1}{a_n}.  $$
From the equality,  $$ P\left(\theta_n\right) =  P\left(\theta_n\right) -  P\left(\theta\right) +  P\left(\theta\right), $$
we get
\begin{equation*}
    \left| P(\theta_n ) \right| \leq \left| P(\theta_n ) - P(\theta ) \right| +\left| P(\theta )\right|.
\end{equation*}
Therefore,

\begin{equation}
    \left| P(\theta ) \right| \geq \left| P(\theta_n ) \right| - \left| P(\theta ) - P(\theta_n ) \right|.
\end{equation}
Put $ \hspace{0.2cm}P = \sum_{k=1}^{d} e_kX^k $, then
\begin{align}
    P(\theta_n ) &= P\left( \frac{p_n}{q_n}\right) = \sum_{k=1}^{d} e_k \frac{p_n^k}{q_n^k} = \frac{1}{q_n^d}\sum_{k=1}^{d} e_kp_n^kq_n^{d-k}.
\end{align}
Notice that
\begin{equation*}
    \sum_{k=1}^{d} e_kp_n^kq_n^{d-k} \neq 0,
\end{equation*}
because, if we assume that $$ \sum_{k=1}^{d} e_kp_n^kq_n^{d-k} = 0, $$\\
then we would have
\begin{align*}
    q_n^d.P\left(\theta_n\right) &= 0, \\
\end{align*}
which implies that
\begin{align*}
P\left(\theta_n\right) &= 0.
\end{align*}
 We also have $$ P\left(\theta\right) = P\left( \lim\limits_{n \rightarrow +\infty} \theta_n\right) = \lim\limits_{n \rightarrow +\infty} P\left(\theta_n\right) = 0, $$\\
which contradicts the fact that $ \theta $ is a transcendental number. Therefore, $$ \left| \sum_{k=1}^{d} e_kp_n^kq_n^{d-k} \right| \geq 1. $$\\
According to equality (9), we get

\begin{equation}
    \left| P\left(\theta_n\right)\right| \geq \frac{1}{q_n^d}.
\end{equation}
On the other hand, according to the mean value theorem applied to $P$, there exists a real number $ F \in ]\theta_n, \theta[ $ or  $]\theta, \theta_n[  $ such that
\begin{equation*}
    P\left( \theta \right) - P\left( \theta_n \right) = P'\left(F\right) \left( \theta - \theta_n \right).
\end{equation*}
From this, we obtain\\
\begin{equation}
    \left| P\left( \theta \right) - P\left( \theta_n \right) \right| = \left| P'\left(F\right) \right| \left| \theta - \theta_n \right|.
\end{equation}
Furthermore, as
\begin{align*}
    P'\left(F\right) &= \sum_{k=1}^{d} k F^{k-1}e_k,
\end{align*}
which implies that
\begin{align*}
     \left| P'\left(F\right) \right| &\leq \sum_{k=1}^{d} k \left| F\right|^{k-1} \left| e_k \right|\\
    &\leq \sum_{k=1}^{d} k \left| e_k \right| \leq H \sum_{k=1}^{d} k\\
    &\leq H\dfrac{d(d+1)}{2}.
\end{align*}
Therefore, the equality (11) becomes
\begin{equation}
    \left| P\left( \theta \right) - P\left( \theta_n \right) \right| < H\dfrac{d(d+1)}{2} \left| \theta - \theta_n \right|.
\end{equation}
According to the relationship (5), we have

\begin{equation*}
    \left| \theta - \theta_n \right| < \frac{1}{q_n^\alpha},
\end{equation*}
then
\begin{equation}
    \left| P\left( \theta \right) - P\left( \theta_n \right) \right| < \dfrac{Hd(d+1)}{2q_n^\alpha}.
\end{equation}
By combining (10) and (13), the relationship (8) becomes
\begin{equation*}
     \left| P\left( \theta \right) \right| > \frac{1}{q_n^d} - \dfrac{Hd(d+1)}{2q_n^\alpha},\hspace{0.06cm} \text{for n sufficiently large.}
\end{equation*}
In order to have $\hspace{0.05cm} \left| P\left( \theta \right) \right| > \dfrac{1}{2q_n^d},$ it suffices to have
\begin{align*}
    \frac{1}{q_n^d} - \dfrac{Hd(d+1)}{2q_n^\alpha} > \frac{1}{2q_n^d},\\
\end{align*}
which is equivalent to
\begin{align*}
    \frac{1}{2q_n^d} &> \dfrac{Hd(d+1)}{2q_n^\alpha}
    \Longleftrightarrow  q_n^{\alpha - d } > Hd(d+1).
\end{align*}
So that, we take $n_1$ the smallest integer such that

\begin{equation}
    q_{n_1-1}^{\alpha - d} < Hd(d+1) < q_{n_1}^{\alpha - d}.
\end{equation}

\begin{Rem}\normalfont
\textrm{The natural number $n_1$ exists because $\lim\limits_{n \rightarrow \infty} q_n^{\alpha - d} = + \infty$, then, we obtain $$q_{n_1}^{\alpha - d} > Hd(d+1).$$}
\end{Rem}
Therefore, we get
\begin{equation}
    \left| p\left( \theta \right) \right| > \frac{1}{2q_{n_1}^d}.
\end{equation}
Using the (ii) from Lemma 3, the relationship (15) becomes
\begin{equation}
    \left| P\left(\theta\right ) \right| > \frac{1}{2q_{n_1}^d} > \frac{1}{2q_{n_1 - 1}^{kd\left(\alpha + 1 \right)}} .
\end{equation}
According to the relationship (14), we have
\begin{align*}
    \frac{1}{q_{n_1-1}^{\alpha - d}} &> \frac{1}{Hd(d+1)},
\end{align*}
then
\begin{align*}
    \frac{1}{q_{n_1-1}^{kd\left(\alpha + 1\right)}} &>\frac{1}{\left(Hd\left(d+1\right)\right)^{\frac{kd\left(\alpha + 1\right)}{\alpha - d}}}.
\end{align*}
So, the relationship (16) becomes
\begin{align*}
     \left| P\left(\theta\right ) \right| &> \dfrac{1}{\left(Hd\left(d+1\right)\right)^{\frac{kd\left(\alpha + 1\right)}{\alpha - d}}},
\end{align*}
which completes the proof of Theorem 2.
$\hspace*{15cm} \Box$
\end{pf}

\begin{Exp}\normalfont
Let $$ \left\{
    \begin{array}{ll}
     a_0 = 0, a_1 = 2,\\
        a_{n+1} = a_n^4, n \geq 1,  \\
        \alpha = 4, k = 2.
    \end{array}
\right. $$

Let $ P \in \mathbb{Z}[X]\textbackslash\left\{ 0 \right\}$ be a quadratic polynomial of height $ H $. By applying Theorem 2, a transcendental measure of $ \theta = \sum_{n = 1}^{\infty} 1/a_n$ is given by

    $$\LARGE{\left| P(\theta) \right| > \dfrac{1}{(6H)^{10}} .}$$

\end{Exp}

