squeeze the floor value of a finite series [duplicate]

elementary-number-theory induction inequality

Solution 1:

$$2\sqrt{k} + \frac {1}{\sqrt{k+1}} = \frac {2\sqrt{k^2+k}+1}{\sqrt{k+1}} < \frac {2\sqrt{k^2+k+\dfrac14}+1}{\sqrt{k+1}} = \frac{2\left(k+\dfrac12\right)+1}{\sqrt{k + 1}}= 2\sqrt{k + 1}.$$

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