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Given that the series $a_n > 0$, and its Harmonic mean and Geometric mean converges to L, does this indecate that $\lim_{n \to \infty}a_n = L$ ?

given $a_n$, the Harmonic mean is defined as : $$ \frac{n}{a_1^{-1}+a_2^{-1}+\ldots+a_n^{-1}} $$ and the Geometric mean is defined as : $$ \sqrt[n]{a_1 \cdot a_2 \cdot\ldots\cdot a_n} $$

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Hint: consider for example the sequence $\;\displaystyle a_n = \begin{cases} \frac{1}{2} \quad \text{if } n = 2^k \text{ for some } k \in \mathbb{N} \\[5px] 1 \quad \text{ otherwise} \end{cases} $

The harmonic and geometric means both converge to $\,L=1\,$ (why?), but $a_n$ itself is not convegent.

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