Carleman's Inequality

Carleman's Inequality states that for nonnegative real numbers $\{a_n\}_{n\ge 1}$, \[\sum_{k=1}^{\infty} (a_1 a_2 \dotsm a_k)^{1/k} < e \sum_{k=1}^{\infty} a_k ,\] unless all the $a_k$ are equal to zero.


Define $c_n = n \left( 1 + \frac{1}{n} \right)^n = \frac{(n+1)^n}{n^{n-1}}$. Then for all positive integers $k$, \[(c_1 \dotsm c_k)^{1/k} = k+1.\] Thus \[\sum_{k=1}^\infty (a_1 \dotsm a_k)^{1/k} = \sum_{k=1}^{\infty}\frac{(c_1a_1 \dotsm c_ka_k)^{1/k}}{(c_1 \dotsm c_k)^{1/k}} = \sum_{k=1}^\infty \frac{(c_1a_1 \dotsm c_ka_k)^{1/k}}{k+1} .\] Now, by AM-GM, \[\sum_{k=1}^\infty \frac{(c_1a_1 \dotsm c_ka_k)^{1/k}}{k+1} \le \sum_{k=1}^{\infty} \sum_{j=1}^k c_ja_j \frac{1}{k(k+1)} = \sum_{j=1}^\infty \sum_{k=j}^{\infty} c_ja_j \frac{1}{k(k+1)} .\] But $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$, so for any integer $j$, \[\sum_{k=j}^\infty \frac{1}{k(k+1)} = \sum_{k=j}^{\infty} \frac{1}{k} - \frac{1}{k+1} = \lim_{N\to \infty} \left( \frac{1}{j} - \frac{1}{N+1} \right) = \frac{1}{j} .\] Therefore \[\sum_{j=1}^{\infty} \sum_{k=j}^{\infty} c_j a_j  \frac{1}{k(k+1)} = \sum_{j=1}^\infty \left( 1 + \frac{1}{j} \right)^j a_k .\] Since $\left( 1 + \frac{1}{j} \right)^j< e$ for all integers $j$, the desired inequality holds. $\blacksquare$

See also


  • Steele, J. M., The Cauchy-Schwarz Master Class, Cambridge University Press, ISBN 0-521-54677-X.
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