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proof of Abel lemma (by expansion)
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(Proof)
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\begin{equation} \sum_{i=0}^n a_ib_i=\sum_{i=0}^{n-1} A_i(b_i-b_{i+1})+A_nb_n, \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.
We consider the expansion of the sum \begin{equation*} \sum_{i=0}^n A_i(b_i-b_{i+1}) \end{equation*}on two different forms, namely:
- On the short way. \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}. \end{equation}
- On the long way.
\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_ib_i-\sum_{i=1}^{n+1} A_{i-1}b_i= \end{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}, \end{equation}where a simplification has been performed. Notice that $A_0=a_0$ . By equating (2), (3), the last two terms cancel, 1 and then, (1) follows. 
Footnotes
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- Without loss of generality, $b_{n+1}$ may be assumed finite. Indeed we don't need $b_{n+1}$ , but the proof is a couple lines larger. It is left as an exercise.
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"proof of Abel lemma (by expansion)" is owned by perucho.
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Cross-references: lines, proof, finite, without loss of generality, terms, sum, complex, real, sequences
This is version 4 of proof of Abel lemma (by expansion), born on 2007-08-12, modified 2007-08-14.
Object id is 9856, canonical name is ProofOfAbelLemmaByExpansion.
Accessed 906 times total.
Classification:
| AMS MSC: | 40A05 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences) |
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Pending Errata and Addenda
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