If \(a_i = 1 - \frac{1}{N_i}\) and \(\sum\limits_{I=0}^k{N_i} = n\), then what is the maximum and minimum values of \(\sum\limits_{I=0}^k{a_i}\)?

Please help, I've tried to solve it but then I got confused. I think I may of found the minimum value to be \(\frac{n - k}{n - k + 1}\) but I'm not sure.

Also \(N_i > 0\) and \(N_i\) is a subset of \(\mathbb{N}\).

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TopNewest\( \displaystyle \sum_{i=0}^{k}a_{i} = k - \displaystyle \sum_{i=0}^{k}\dfrac{1}{N_{i}} \)

Using AM-GM-HM inequality,

\( \dfrac{\displaystyle \sum_{i=0}^{k}N_{i}}{k} \ge \dfrac{k}{\displaystyle \sum_{i=0}^{k}\dfrac{1}{N_{i}}} \)

\( \displaystyle \sum_{i=0}^{k} \dfrac{1}{N_{i}} \ge \dfrac{k^{2}}{n} \)

\( -\displaystyle \sum_{i=0}^{k} \dfrac{1}{N_{i}} \le - \dfrac{k^{2}}{n} \)

\( \displaystyle \sum_{i=0}^{k} a_{i} \le k - \dfrac{k^{2}}{n} \)

This is the maximum value of the expression, I am not sure about the minimum. – Vighnesh Shenoy · 9 months, 1 week ago

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– Harry Obey · 9 months ago

Thanks so much, I have been trying to solve this problem for about 3 days. I heard that it may be possible to find the minimum using Lagrange multipliers but I'm not sure.Log in to reply