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show that $$\sum_{i=1}^{n}\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{i}}{i^2}\le\sqrt{2n-1}$$

My try: $$x\in(n-1,n)\Longrightarrow \sqrt{x}>\sqrt{n-1}$$ so $$\int_{n-1}^{n}\sqrt{x}dx>\sqrt{n-1}$$ so $$\dfrac{2}{3}n^{\frac{3}{2}}=\int_{0}^{n}\sqrt{x}dx>\sqrt{n-1}+\sqrt{n-2}+\cdots+1$$ so $$1+\sqrt{2}+\cdots+\sqrt{n}<\sqrt{n}+\dfrac{2}{3}n^{\frac{3}{2}}$$ But this not use.Thank you

math110
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1 Answers1

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Try induction. $P(1)$ is obvious.

$P(n) \Rightarrow P(n_+1)$.

$$\sum_{i=1}^{n+1}\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{i}}{i^2}\le\sqrt{2n-1} +\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1}}{(n+1)^2}$$

To complete the proof we need to show

$$\sqrt{2n-1} +\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1}}{(n+1)^2} \leq \sqrt{2n+1} \Leftrightarrow \\ \dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1}}{(n+1)^2} \leq \sqrt{2n+1}-\sqrt{2n-1} \Leftrightarrow \\ \sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1} \leq (n+1)^2[ \sqrt{2n+1}-\sqrt{2n-1}] = \frac{2(n+1)^2}{\sqrt{2n-1}+\sqrt{2n+1}} $$

Now, by your inequality we have

$$\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1} \leq \sqrt{n+1}+\frac{2}{3}(n+1)^{\frac{3}{2}}$$

and the inequality

$$\sqrt{n+1}+\frac{2}{3}(n+1)^{\frac{3}{2}} \leq \frac{2(n+1)^2}{2\sqrt{2n+1}} \leq \frac{2(n+1)^2}{\sqrt{2n-1}+\sqrt{2n+1}} $$

is easy to prove and completes the solution.

N. S.
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  • It's very nice.Thank you+1 – math110 Nov 04 '13 at 16:02
  • Shouldn't it be $\sqrt{2n+1}-\sqrt{2n-1} = \frac{2}{\sqrt{2n-1}+\sqrt{2n+1}}$? – Alex Nov 04 '13 at 16:31
  • @Alex Ty, fixed it. The last inequality should still be true since $\frac{2 \sqrt{2}}{3} <1$, but it might fail for small $n$. This means that it needs to be checked manually up to a hopefully reasonable value :) – N. S. Nov 04 '13 at 18:58