Difference between revisions of "Abel inequality"
From Encyclopedia of Mathematics
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− | An estimate for the sum of products of two numbers. If sets of numbers | + | An estimate for the sum of products of two numbers. If sets of numbers $(a_k)$ and $(n_k)$ are given such that the absolute values of all sums $B_k = b_1 + \cdots + b_k$, $k=1,2,\ldots$, are bounded by a number $B$, i.e. $|B_k| \le B$, and if either $a_i \ge a_{i+1}$ or $a_i \le a_{i+1}$, $i = 1,2,\ldots,n-1$, then |
+ | $$ | ||
+ | \left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B(|a_1| + 2|a_n|) | ||
+ | $$ | ||
− | + | If the $a_k$ are non-increasing and non-negative, one has the simpler estimate: | |
+ | $$ | ||
+ | \left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B a_1 \ . | ||
+ | $$ | ||
− | |||
− | + | Abel's inequality is proved by means of the [[Abel transformation]]. | |
− | + | {{TEX|done}} |
Revision as of 12:11, 9 November 2014
An estimate for the sum of products of two numbers. If sets of numbers $(a_k)$ and $(n_k)$ are given such that the absolute values of all sums $B_k = b_1 + \cdots + b_k$, $k=1,2,\ldots$, are bounded by a number $B$, i.e. $|B_k| \le B$, and if either $a_i \ge a_{i+1}$ or $a_i \le a_{i+1}$, $i = 1,2,\ldots,n-1$, then $$ \left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B(|a_1| + 2|a_n|) $$
If the $a_k$ are non-increasing and non-negative, one has the simpler estimate: $$ \left\vert{ \sum_{k=1}^n a_k b_k }\right\vert \le B a_1 \ . $$
Abel's inequality is proved by means of the Abel transformation.
How to Cite This Entry:
Abel inequality. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Abel_inequality&oldid=18342
Abel inequality. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Abel_inequality&oldid=18342
This article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article