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How to Prove Abel's Formula ?

Abel's Formula is -

Let $(a_n)$ and $(b_n)$ be sequences of real numbers. The Abel’s formula reads, for $p \geq 2$,

$\sum_{n=1}^{p} a_nb_n $ = $\sum_{k=1}^{p-1} (a_k - a_{k+1}) (\sum_{l=1}^{k} b_l)$ + $ a_p \sum_{l=1}^{p} b_l $

or can be written as

$\sum_{n=1}^{p} a_nb_n $ = $ (a_1 - a_2)b_1 + (a_2 - a_3)(b_1 + b_2)+(a_3 - a_4)(b_1 + b_2 + b_3) + ··· + (a_{p-1} - a_p)(b_1 + ··· + b_{p-1}) + a_p(b_1 + ··· + b_p)$

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No induction is required to prove Abel's formula:

Set $\,B_n=\displaystyle\sum_{l=1}^{n}b_l\,$; for all $n>1$, we have $b_n=B_n-B_{n-1}$, and $b_1=B_1$. With these notations, we can write: \begin{align*} \sum_{n=1}^p a_nb_n &= a_1 B_1+\sum_{n=2}^p a_n(B_n-B_{n-1})=\sum_{n=1}^p a_n B_n -\sum_{k=1}^{p-1} a_{k+1} B_k\\ &= a_pB_p +\sum_{k=1}^{p-1}(a_k - a_{k+1}) B_k \end{align*}

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  • $\begingroup$ In the equation How you changed from n to k ? Is there any relation between k and n ? $\endgroup$ – pratik Apr 8 '15 at 15:25
  • $\begingroup$ I renamed $n-1$ to $k$ in first line, and $n$ to $k$ in second line (no problem: it's a dummy variable). $\endgroup$ – Bernard Apr 8 '15 at 15:32
  • $\begingroup$ Dont you think it must be $\sum_{k=1}^{p-1}$ in the second part of the equation instead of k = 0 ? $\endgroup$ – pratik Apr 8 '15 at 20:03
  • $\begingroup$ You're perfectly right. Thanks for pointing it. Actually, I adapted on the fly a demo in which the indices started at 0 and I got messed up! $\endgroup$ – Bernard Apr 8 '15 at 20:19
  • $\begingroup$ Induction is required to prove that summation and addition/subtraction commute. $\endgroup$ – user21820 Apr 10 '15 at 13:33
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Hint: When you look at the R.H.S. and calculate the difference at the $p$-th induction step and $(p+1)$-th induction step, it is \begin{align} & \big[\sum_{k=1}^{p} (a_k - a_{k+1})\sum_{l=1}^k b_l + a_{p+1} \sum_{l=1}^{p+1} b_l\big] - \big[\sum_{k=1}^{p-1} (a_k - a_{k+1})\sum_{l=1}^k b_l + a_{p} \sum_{l=1}^p b_l\big]\\ = & \big[\sum_{k=1}^{p} (a_k - a_{k+1})\sum_{l=1}^k b_l - \sum_{k=1}^{p-1} (a_k - a_{k+1})\sum_{l=1}^k b_l \big] + \big[(a_{p+1}b_{p+1} + a_{p+1} \sum_{l=1}^p b_l) - a_p\sum_{l=1}^pb_l\big]\\ = & (a_p - a_{p+1})\sum_{l=1}^p b_l + a_{p+1}b_{p+1} + (a_{p+1} - a_p)\sum_{l=1}^pb_l\\ = & a_{p+1}b_{p+1} \end{align}

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  • $\begingroup$ Do I need to prove whole Proof by induction only ? Or I am suppose to prove Abel's theorem by another method and then am I suppose to do it by induction as well ? $\endgroup$ – pratik Apr 8 '15 at 14:51
  • $\begingroup$ I don't think the above induction step need any other special methods. It just rearranges the terms. But I usually regard it as a corollary of the summation by part formula. $\endgroup$ – Empiricist Apr 8 '15 at 14:58
  • $\begingroup$ So It can Be proved by corollary of the summation by part formula. ? $\endgroup$ – pratik Apr 8 '15 at 15:17

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