Difference between revisions of "Carleman's Inequality"
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<cmath> \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)} . </cmath> | <cmath> \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)} . </cmath> | ||
But <math>\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}</math>, so for any integer <math>j</math>, | But <math>\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}</math>, so for any integer <math>j</math>, | ||
| − | <cmath> \sum_{k=j}^\infty \frac{1}{k(k+1)} = \sum_{k=j}^{\infty} \frac{1}{ | + | <cmath> \sum_{k=j}^\infty \frac{1}{k(k+1)} = \sum_{k=j}^{\infty} \frac{1}{k} - \frac{1}{k+1} = \lim_{N\to \infty} \left( \frac{1}{j} - \frac{1}{N+1} \right) = \frac{1}{j} . </cmath> |
Therefore | Therefore | ||
| − | <cmath> \sum_{j=1}^{\infty} \sum_{k=j}^{\infty} | + | <cmath> \sum_{j=1}^{\infty} \sum_{k=j}^{\infty} c_j a_j \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> | ||
Revision as of 00:56, 3 September 2013
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 
,
![\[\sum_{k=j}^\infty \frac{1}{k(k+1)} = \sum_{k=j}^{\infty} \frac{1}{k} - \frac{1}{k+1} = \lim_{N\to \infty} \left( \frac{1}{j} - \frac{1}{N+1} \right) = \frac{1}{j} .\]](http://latex.artofproblemsolving.com/e/9/9/e99b0f31a36d3e338b80329f973d10e964898f85.png)
Therefore
![\[\sum_{j=1}^{\infty} \sum_{k=j}^{\infty} c_j a_j \frac{1}{k(k+1)} = \sum_{j=1}^\infty \left( 1 + \frac{1}{j} \right)^j a_k .\]](http://latex.artofproblemsolving.com/e/4/9/e4939e2cf4dc1817802745b1b97de9f670d8f240.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.
