# 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.

## Proof

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. $\blacksquare$