| Version 2 |
Version 1 |
| \section{Abel lemma} |
\section{Abel lemma} |
| \begin{equation} |
\begin{equation} |
| \sum_{i=0}^n a_ib_i=\sum_{i=0}^{n-1} A_i(b_i-b_{i+1})+A_nb_n, |
\sum_{i=0}^n a_ib_i=\sum_{i=0}^{n-1} A_i(b_i-b_{i+1})+A_nb_n, |
| \end{equation} |
\end{equation} |
| where, $A_i=\sum_{k=0}^i a_k$. Sequences $\{a_i\}$, $\{b_i\}$, $i=0,\dots, n$, are real or complex one. |
where, $A_i=\sum_{k=0}^i a_k$. Sequences $\{a_i\}$, $\{b_i\}$, $i=0,\dots, n$, are real or complex one. |
| \section{Proof} |
\section{Proof} |
| We consider the expansion of the sum |
We consider the expansion of the sum |
| \begin{equation*} |
\begin{equation*} |
| \sum_{i=0}^n A_i(b_i-b_{i+1}) |
\sum_{i=0}^n A_i(b_i-b_{i+1}) |
| \end{equation*} |
\end{equation*} |
| on two different forms, namely: |
on two different forms, namely: |
| \begin{enumerate} |
\begin{enumerate} |
| \item On the short way. |
\item On the short way. |
| \begin{equation} |
\begin{equation} |
| \sum_{i=0}^n A_i(b_i-b_{i+1})=\sum_{i=0}^{n-1} A_i(b_i-b_{i+1})+A_nb_n-A_nb_{n+1}. |
\sum_{i=0}^n A_i(b_i-b_{i+1})=\sum_{i=0}^{n-1} A_i(b_i-b_{i+1})+A_nb_n-A_nb_{n+1}. |
| \end{equation} |
\end{equation} |
| \item On the long way. |
\item On the long way. |
| \end{enumerate} |
\end{enumerate} |
| \begin{equation*} |
\begin{equation*} |
| \sum_{i=0}^n A_i(b_i-b_{i+1})=\sum_{i=0}^n A_ib_i-\sum_{i=0}^n A_ib_{i+1}= |
\sum_{i=0}^n A_i(b_i-b_{i+1})=\sum_{i=0}^n A_ib_i-\sum_{i=0}^n A_ib_{i+1}= |
| \sum_{i=0}^n A_ib_i-\sum_{i=1}^{n+1} A_{i-1}b_i= |
\sum_{i=0}^n A_ib_i-\sum_{i=1}^{n+1} A_{i-1}b_i= |
| \end{equation*} |
\end{equation*} |
| \begin{equation} |
\begin{equation} |
| A_0 b_0+\sum_{i=1}^n (A_{i-1}+a_i)b_i-\sum_{i=1}^n A_{i-1}b_i-A_nb_{n+1}=\sum_{i=0}^n a_ib_i-A_nb_{n+1}, |
A_0 b_0+\sum_{i=1}^n (A_{i-1}+a_i)b_i-\sum_{i=1}^n A_{i-1}b_i-A_nb_{n+1}=\sum_{i=0}^n a_ib_i-A_nb_{n+1}, |
| \end{equation} |
\end{equation} |
|
where a simplification has been performed. Notice that $A_0=a_0$. By equating (2), (3), the last two terms cancel, and then, (1) follows. $\Box$
|
where a simplification has been performed. Notice that $A_0=a_0$. By comparing (2), (3), the last two terms cancel, and then, (1) follows. $\Box$
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