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How can I find a tight upper bound for the following expression:

$\sum\limits_{i=1}^{k} a_i \sum\limits_{j = 1}^{i} \frac{1}{b_j} = a_1 \frac{1}{b_1} + a_2 (\frac{1}{b_1} + \frac{1}{b_2}) + \dots + a_k (\frac{1}{b_1} + \frac{1}{b_2} \dots \frac{1}{b_k})$

where, $b_1 \geq b_2 \geq \dots \geq b_k$, $\forall i ~ a_i, b_i > \mathbb{Z}^{+}$?

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Since $b_i\geqslant b_k$, $ \frac{1}{b_j}\leqslant \frac{1}{b_k}$ hence $$\sum\limits_{i=1}^{k} a_i \sum\limits_{j = 1}^{i} \frac{1}{b_j} \leqslant \frac{1}{b_k} \sum\limits_{i=1}^{k} ia_i.$$ This is tight if all the $b_i$ are equal.

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  • $\begingroup$ Yeah, if all $b_i$'s are equal, this is good. But, It would be great if we can do anything better considering all $b_i$'s are not equal? $\endgroup$ – jayesh May 7 '17 at 17:44

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