Difference between revisions of "Carleman's Inequality"
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<cmath> \sum_{j=1}^{\infty} \sum_{k=j}^{\infty} c_k a_k = \sum_{j=1}^{\infty} j \left(1 + \frac{1}{j} \right)^j a_j \sum_{k=j}^{\infty} \frac{1}{k(k+1)} = \sum_{j=1}^\infty \left( 1 + \frac{1}{j} \right)^j a_k . </cmath> | <cmath> \sum_{j=1}^{\infty} \sum_{k=j}^{\infty} c_k a_k = \sum_{j=1}^{\infty} j \left(1 + \frac{1}{j} \right)^j a_j \sum_{k=j}^{\infty} \frac{1}{k(k+1)} = \sum_{j=1}^\infty \left( 1 + \frac{1}{j} \right)^j a_k . </cmath> | ||
Since <math>\left( 1 + \frac{1}{j} \right)^j< e</math> for all integers <math>j</math>, the desired inequality holds. <math>\blacksquare</math> | Since <math>\left( 1 + \frac{1}{j} \right)^j< e</math> for all integers <math>j</math>, the desired inequality holds. <math>\blacksquare</math> | ||
| + | |||
| + | == See also == | ||
| + | *[[Inequality]] | ||
| + | *[[Nonnegative]] | ||
== References == | == References == | ||
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[[Category:Inequality]] | [[Category:Inequality]] | ||
| + | [[Category:Theorems]] | ||
Revision as of 21:44, 1 January 2008
Carleman's Inequality states that for nonnegative real numbers 
,
![\[\sum_{k=1}^{\infty} (a_1 a_2 \dotsm a_k)^{1/k} < e \sum_{k=1}^{\infty} a_k ,\]](http://latex.artofproblemsolving.com/d/1/1/d111396ef9d06b06ad81fd4606ea2608b8ba13bd.png)
unless all the 
are equal to zero.
Proof
Define 
. Then for all positive integers 
,
![\[(c_1 \dotsm c_k)^{1/k} = k+1.\]](http://latex.artofproblemsolving.com/3/4/0/3403776f9ba18d681ba17e0ac7f18c0e7f96d3de.png)
Thus
![\[\sum_{k=1}^\infty (a_1 \dotsm a_k)^{1/k} = \sum_{k=1}^{\infty}\frac{(c_1a_1 \dotsm c_ka_k)^{1/k}}{(c_1 \dotsm c_k)^{1/k}} = \sum_{k=1}^\infty \frac{(c_1a_1 \dotsm c_ka_k)^{1/k}}{k+1} .\]](http://latex.artofproblemsolving.com/f/7/e/f7e13ffcd2b1f6871eeb58fc32a73f63ee69b436.png)
Now, by AM-GM,
![\[\sum_{k=1}^\infty \frac{(c_1a_1 \dotsm c_ka_k)^{1/k}}{k+1} \le \sum_{k=1}^{\infty} \sum_{j=1}^k c_ja_j \frac{1}{k(k+1)} = \sum_{j=1}^\infty \sum_{k=j}^{\infty} c_ja_j \frac{1}{k(k+1)} .\]](http://latex.artofproblemsolving.com/d/8/5/d85af8e1bc46cb4b28aafc222347eb8ce6a156de.png)
But 
, so for any integer 
,
![\[\sum_{k=j}^\infty \frac{1}{k(k+1)} = \sum_{k=j}^{\infty} \frac{1}{j} - \frac{1}{j+1} = \lim_{j\to \infty} \left( \frac{1}{k} - \frac{1}{j+1} \right) = \frac{1}{k} .\]](http://latex.artofproblemsolving.com/3/7/7/37790d55db9efb9f65dc9c4166fc9d28bf342cf3.png)
Therefore
![\[\sum_{j=1}^{\infty} \sum_{k=j}^{\infty} c_k a_k = \sum_{j=1}^{\infty} j \left(1 + \frac{1}{j} \right)^j a_j \sum_{k=j}^{\infty} \frac{1}{k(k+1)} = \sum_{j=1}^\infty \left( 1 + \frac{1}{j} \right)^j a_k .\]](http://latex.artofproblemsolving.com/6/4/f/64f72d50500c195a1fae47f05b99395102f315db.png)
Since 
for all integers , the desired inequality holds. 
See also
References
- Steele, J. M., The Cauchy-Schwarz Master Class, Cambridge University Press, ISBN 0-521-54677-X.
