Prove that $$\lim\limits_{n \to \infty}\frac1n \sum\limits_{k=1}^nn^{1/k}=2$$
How to estimate the sum? I tried to use Stolz's theorem but it is still a similar summation.
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1This is wrong: $$\sum_{k=1}^{n}n^{1/k}\ge n^{1/1}+\sum_{k=2}^{n}1=2n-1.$$ – Sep 29 '17 at 07:07
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@ProfessorVector Oh, sorry. I have corrected the range of summation, which begins at 2. – Kirby Lee Sep 29 '17 at 07:11
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It's still wrong in the title of your question. – Sep 29 '17 at 07:12
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@ProfessorVector Either slow down... or take things in your own hands by correcting the typos. :-) – Did Sep 29 '17 at 07:12
1 Answers
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Hint:
Using the arithmetic-geometric mean inequality
$$1 \leqslant n^{1/k} = (\underbrace{1 \cdot 1 \cdots 1}_{\text{k-2 times}} \sqrt{n} \sqrt{n})^{1/k} \leqslant \frac{1}{k} (k-2 + 2\sqrt{n}) = 1 + \frac{2(\sqrt{n}-1)}{k}$$
Sum on $2\leqslant k\leqslant n$ and apply the squeeze theorem.
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How does adding and dividing by $n$ leads to limit $2$? Shouldn't it lead to $1$? – Paramanand Singh Sep 29 '17 at 07:18
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@Paramanand Singh: $n- 1 \leqslant \sum_{k=2}^n n^{1/k} \leqslant \ldots$ and $2n- 1 \leqslant \sum_{k=1}^n n^{1/k} \leqslant \ldots$ – RRL Sep 29 '17 at 07:20
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+1 already for the very smart use of AM GM. I wonder how much powerful the simple inequality is. – Paramanand Singh Sep 29 '17 at 07:20