With $n\in\mathbb{N}\setminus\{0\}$ how can I prove that $$ 1) \quad\sqrt[n]{n} < 1 + \dfrac{1}{\sqrt{n}} \\ 2) \quad \sqrt[n]{n} < 1 + \dfrac{2}{\sqrt{n}} $$?
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Hint: $$\sqrt[n]n<1+2n^{-1/2}\Longleftrightarrow n<\sum_{k=0}^n\binom{n}{k} 2^kn^{-k/2}=1+2\sqrt n+2(n-1)+\ldots$$ |
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