# proof of Abel’s lemma (by induction)

*Proof.* The proof is by induction^{}. However, let us
first recall that sum on the right side is a
piece-wise defined function of the upper limit^{} $N-1$.
In other words, if the upper limit is below the lower
limit $0$, the sum is identically set to zero.
Otherwise, it is an ordinary sum.
We therefore need to manually check the first two cases.
For the trivial case $N=0$, both sides equal to ${a}_{0}{b}_{0}$.
Also, for $N=1$ (when the sum is a normal sum), it is easy to verify that
both sides simplify to ${a}_{0}{b}_{0}+{a}_{1}{b}_{1}$.
Then, for the induction step, suppose that the
claim holds for some $N\ge 1$. For $N+1$, we then have

$\sum _{i=0}^{N+1}}{a}_{i}{b}_{i$ | $=$ | $\sum _{i=0}^{N}}{a}_{i}{b}_{i}+{a}_{N+1}{b}_{N+1$ | ||

$=$ | $\sum _{i=0}^{N-1}}{A}_{i}({b}_{i}-{b}_{i+1})+{A}_{N}{b}_{N}+{a}_{N+1}{b}_{N+1$ | |||

$=$ | $\sum _{i=0}^{N}}{A}_{i}({b}_{i}-{b}_{i+1})-{A}_{N}({b}_{N}-{b}_{N+1})+{A}_{N}{b}_{N}+{a}_{N+1}{b}_{N+1}.$ |

Since $-{A}_{N}({b}_{N}-{b}_{N+1})+{A}_{N}{b}_{N}+{a}_{N+1}{b}_{N+1}={A}_{N+1}{b}_{N+1}$, the claim follows. $\mathrm{\square}$.

Title | proof of Abel’s lemma (by induction) |
---|---|

Canonical name | ProofOfAbelsLemmabyInduction |

Date of creation | 2013-03-22 13:38:04 |

Last modified on | 2013-03-22 13:38:04 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 9 |

Author | mathcam (2727) |

Entry type | Proof |

Classification | msc 40A05 |