Positive-definiteness of matrix with $|a_{ij}| \leq \frac{1}{ij}$

Let $A$ be a $N\times N$-matrix with elements $$ a_{ii}=1 \quad\text{and}\quad a_{ij} = \frac{1}{ij} \quad\text{for}~ i\neq j. $$ Then $A$ is positive-definite, as can be easily seen from $$ x^T A x = \sum_i x_i^2 + \sum_{i \neq j} \frac{x_i x_j}{ij} \geq \sum_i \frac{x_i^2}{i^2} + \sum_{i \neq j} \frac{x_i x_j}{ij} = \left(\sum_i \frac{x_i}{i}\right)^2 \geq 0. $$

Assume now that $A$ is a real symmetric $N\times N$-matrix with elements $$ a_{ii}=1 \quad\text{and}\quad |a_{ij}| \leq \frac{1}{ij} \quad\text{for}~ i\neq j. $$

Is it possible to show that $A$ is also positive-definite (or positive-semidefinite)?

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