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For any positive integer $n$ prove by induction that:
$$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2.$$

The author says that it is sufficient to prove that
$$ \frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\dots+\frac{1}{(n+1)\sqrt{n}}<2-\frac{2}{\sqrt{n+1}}.$$
Why? Where this $\frac{2}{\sqrt{n+1}}$ term come from?

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  • $\begingroup$ By transitivity of $<$ ? ... $\endgroup$ – Suzet Jun 25 '18 at 7:32
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    $\begingroup$ Hint: $\frac{1}{(n+1)\sqrt{n}} \le \frac{2}{(\sqrt{n+1}+\sqrt{n})\sqrt{n}\sqrt{n+1}} = \cdots$ $\endgroup$ – achille hui Jun 25 '18 at 7:43
  • $\begingroup$ I think you only get such a nice value for $(An+B)^{-1/2}$ because of the specific numerical coefficients and constants in the summation. If the problem had been $\sum ((k+\pi)\sqrt{4k + \sqrt 7})^{-1}$ or something like that, there wouldn't be such a pretty magic term like that at all. This smells like a solution pretending to be a problem. $\endgroup$ – DanielV Jun 25 '18 at 10:43
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The stronger inequality is easier to be proved by using induction than the original one. This is another example: in order to prove
$$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} < 2$$ show that $$\dfrac{1}{1^2}+ \dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2} \le 2-\dfrac{1}{n}.$$ See Induction on inequalities: $\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\ldots+\frac1{n^2}<2$?

In our case, at the inductive step, it suffices to show that $$2-\frac{2}{\sqrt{n+1}}+\frac{1}{(n+2)\sqrt{n+1}}<2-\frac{2}{\sqrt{n+2}}.$$ Can you take it from here?

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  • $\begingroup$ I don't think he is asking how to use the given value. I think he is asking where this magic term comes from. Even if you somehow manage to guess the ansatz $(A+Bn)^{-1/2}$ it isn't obvious how to get $A=1/4,B=1/4$. $\endgroup$ – DanielV Jun 25 '18 at 10:11
  • $\begingroup$ @DanielV I agree with you. I just tried to answer to the "Why?" $\endgroup$ – Robert Z Jun 25 '18 at 10:25
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The idea of the author is that if you "stop" the sum at the $n$'th term you get this artificial bound of $2-$something. Then you can show by induction that this holds for every $n$. Having proven this, the assertion follows by taking the limit as $n\rightarrow \infty$ since the bound will always be less than $2$.

Long story short, the author creates a bound which is provable by induction, that's where the $\frac{2}{\sqrt{n+1}}$ comes from

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