Let $0 < p < 1/2$. How can I show the following lower bound?

$$ sum_{i=1}^n 1-expleft(- frac{np^i}{i}right) = Omega(log n) $$

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# How to show that $sum_{i=1}^n 1-e^{-n p^i/i} = Omega(log n)$ for $0 < p < 1/2$?

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Let $0 < p < 1/2$. How can I show the following lower bound?

$$ sum_{i=1}^n 1-expleft(- frac{np^i}{i}right) = Omega(log n) $$

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