Inequalities: $\sqrt[n]{n} < 1 + \dfrac{1}{\sqrt{n}}$

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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What is your question? –  Harald Hanche-Olsen Jan 28 '13 at 13:28
The question has been posted here too-artofproblemsolving.com/Forum/viewtopic.php?f=51&t=518373 –  user54807 Jan 28 '13 at 13:57
you can help me? –  thannhantrung211 Jan 28 '13 at 14:13
Of course we can. The question is whether we should. (Solution posted at AOPS.) –  Did Jan 28 '13 at 14:31

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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thank you but you can other solutions? –  thannhantrung211 Jan 28 '13 at 14:48
I don't understand you. –  DonAntonio Jan 28 '13 at 16:04
i want to have other solutions! :) –  thannhantrung211 Jan 28 '13 at 16:16
Good for you...so what? You could at least show some appreciation for one solution... –  DonAntonio Jan 28 '13 at 17:10