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I have the following inequality to prove with induction:

$$P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots\frac{1}{\sqrt{n}}>2-\frac{2}{n}, \forall n\in \mathbb{\:N}^*$$

I tried to prove $P(n+1)$:

Let $S = \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\Rightarrow$

$$P(n+1):S>2-\frac{2}{n+1}$$

I got into this point and I think it is all wrong, but I'll write it here also:

$$S>\frac{2(n+1) + n\sqrt{n+1}}{n(n+1)}$$

and I don't know what to do next... Could anybody help me, please? I would also like to know if there's any other smarter way of solving this kind of exercises.

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    $\begingroup$ It might be easier to look at $P(1),P(2),P(3)$ and then wonder whether there is an error in the question $\endgroup$ – Henry Dec 17 '14 at 7:54
  • $\begingroup$ There is not any error in the question. I might have gone on the wrong path. Because $P(1): 1>0 \Rightarrow true$, $P(2):2+\sqrt{2}>2 \Rightarrow true$ and $P(3): 6+3\sqrt{2}+2\sqrt{3}>8 \Rightarrow true$ $\endgroup$ – Victor Dec 17 '14 at 7:59
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    $\begingroup$ But $\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}} \gt 2 \gt 2-\dfrac2n$ for all positive $n$ $\endgroup$ – Henry Dec 17 '14 at 8:02
  • $\begingroup$ Assume $P(n-1)$ is true. then: $$ \sum_{k=1}^{n-1}{1\over\sqrt{k}} < 2-{{2}\over{n-1}} \\ \sum_{k=1}^{n}{1\over\sqrt{k}} \,<\, 2+{{1}\over{\sqrt{n}}}-{{2}\over{n-1}} \\ $$ Note that: $$ {2\over{n}}<{{2}\over{n-1}}-{{1}\over{\sqrt{n}}} \\ $$ $\endgroup$ – SomeStrangeUser Dec 17 '14 at 8:02
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    $\begingroup$ A more interesting question might be showing the sum is between $\sqrt{n}$ and $2\sqrt{n}$ $\endgroup$ – Henry Dec 17 '14 at 8:21
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Initial comment: Begin by noting that, for all $n\geq 1$, we have that $$ n(\sqrt{n}-2)+2>0\Longleftrightarrow n\sqrt{n}-2n+2>0\Longleftrightarrow \color{red}{\sqrt{n}>2-\frac{2}{n}}.\tag{1} $$ Thus, it suffices for us to prove the proposition $P(n)$ for all $n\geq 1$ where $$ P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\geq \sqrt{n}.\tag{2} $$ If we can prove $(2)$, then we will have proven $$ \color{blue}{\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}}\color{red}{\geq\sqrt{n}}\color{blue}{> 2-\frac{2}{n}}, $$ as desired. I'm sure you can handle the proof of $(1)$ quite easily.

Claim: For $n\geq 1$, let $P(n)$ denote the statement $$ P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\geq \sqrt{n}. $$

Base step: $P(1)$ holds since $1\geq\sqrt{1}$ is true.

Before induction step: Consider the following inequality for any $x\geq 1$: $$ \sqrt{x}+\frac{1}{\sqrt{x+1}}>\sqrt{x+1}\tag{3}. $$ Briefly, observe that for $x\geq 1, \sqrt{x(x+1)}>x$; thus, $\sqrt{x(x+1)}+1>x+1$. Dividing by $\sqrt{x+1}$ proves $(3)$. The purpose of $(3)$ is to streamline the calculations below in the inductive step.

Inductive step: Fix some $k\geq 1$ and suppose that $P(k)$ is true. Then \begin{align} \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} &\geq \sqrt{k}+\frac{1}{\sqrt{k+1}}\tag{by $P(k)$}\\[1em] &> \sqrt{k+1},\tag{by $(3)$} \end{align} which shows that $S(k+1)$ follows. This concludes the inductive step.

Thus, for all $n\geq 1, P(n)$ is true. $\blacksquare$

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