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Let $k,n \in \mathbb{Z}^+$ with $n > 1$. Prove that $$\frac{1}{kn} + \frac{1}{kn + 1} + \dotsb + \frac{1}{kn + n - 1} > n \left(\sqrt[n]{\frac{k+1}{k}} - 1 \right)$$

I roughly observe that AM-GM can be used so I tried that $$\sum_{k = 0}^{n-1}\frac{1}{kn + j} + n = \sum_{k = 0}^{n-1}\frac{kn^2 + jn + 1}{kn + j}$$

but I am stuck how to proceed.

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    $\begingroup$ I wonder if the second fraction should be $\frac{1}{kn + 1}$. $\endgroup$
    – Martin R
    Commented May 7, 2021 at 5:43

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Your idea of using AM-GM inequality is right, but unfortunately your equation is wrong. It should be $$\left(\sum_{j=0}^{n-1}\frac{1}{kn+j}\right)+n=\sum_{j=0}^{n-1}\left(\frac{1}{kn+j}+1\right)=\sum_{j=0}^{n-1}\frac{kn+j+1}{kn+j}$$ Then it should be easier for you to proceed from here, as we observe that $$\prod_{j=0}^{n-1}\frac{kn+j+1}{kn+j}=\frac{kn+n}{kn}=\frac{k+1}k$$

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