Carleman Inequality

Statement

If a positive series \(\sum_{n=1}^{\infty} b_n\) converges, \(\sum_{n=1}^{\infty} \sqrt[n]{b_1b_2\cdots b_n}\) converges, and \(\sum_{n=1}^{\infty} \sqrt[n]{b_1b_2\cdots b_n} \leq e \sum_{n=1}^{\infty} b_n\).

Proof

For any \(n \in Z^+\), let’s prove that \(\sum_{n=1}^{k} \sqrt[k]{b_1b_2\cdots b_n}\) is upper bounded.

\[\sqrt[k]{b_1b_2\cdots b_k} = \sqrt[k]{ \frac{\prod_{i=1}^k i b_i}{k!}}.\]

Then we apply (k+1)/e^k<k! and have

\[\sqrt[k]{ \frac{\prod_{i=1}^n i b_i}{k!}} < \frac{e}{k+1} \sqrt[k]{\prod_{i=1}^k ib_i}.\]

Now use Cauchy-Schwarz inequality and have

\[\sqrt[k]{\prod_{i=1}^k ib_i} \leq \frac{1}{k} \sum_{i=1}^k ib_i.\]

Therefore,

\[\sum_{k=1}^{n} \sqrt[n]{b_1b_2\cdots b_n} < \sum_{k=1}^n \frac{e}{k(k+1)}\sum_{i=1}^k ib_i = \sum_{k=1}^n \sum_{i=1}^k \frac{e}{k(k+1)}ib_i = e\sum_{k=1}^n \sum_{i=1}^k (\frac{1}{k} - \frac{1}{k+1}) ib_i < e\sum_{k=1}^n \sum_{i=1}^k \frac{1}{k} ib_i = e \sum_{i=1}^n b_i \leq e \sum_{i=1}^\infty b_i.\]