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I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$

Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$

Step: assume holds for $n_0$.

$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n + 1}}$. But I do not know what to do next? How this can be proved?

Américo Tavares
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Bob
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5 Answers5

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You assume it holds for $n$, i.e. $$ 2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}}. $$ You now want to prove that it holds for $n+1$, i.e. $$ 2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}. $$ Subtracting the first inequality from the second, you only have to show that $$ 2(\sqrt{n+2}-\sqrt{n+1}) < \frac{1}{\sqrt{n+1}}. $$ Rearrangement yields $$ \frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+1}} = \frac{\sqrt{n+2}}{\sqrt{n+1}} - 1 < \frac{1}{2}, $$ so $$ \frac{\sqrt{n+2}}{\sqrt{n+1}} < \frac{3}{2}. $$ Squaring both sides gives us $$ \frac{n+2}{n+1} < \frac{9}{4}, $$ after which cross multiplication results in $$ 4n+8 < 9n+9, $$ which is obvious for all $n \geq 1$.

Tim Vermeulen
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For induction step, it's enough to prove $\frac{1}{\sqrt{n+1}}>2(\sqrt{n+2}-\sqrt{n+1})$. $$2(\sqrt{n+2}-\sqrt{n+1})=\frac{2}{\sqrt{n+2}+\sqrt{n+1}}<\frac{2}{2\sqrt{n+1}}=\frac{1}{\sqrt{n+1}}$$

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\begin{align} \frac1{\sqrt{n+1}}+2\sqrt{n+1}-2&=\frac{1+2(n+1)}{\sqrt{n+1}}-2\\ &=\frac{\sqrt{4n^2+12n+9}}{\sqrt{n+1}}-2\\ &>\frac{\sqrt{4n^2+12n+8}}{\sqrt{n+1}}-2\\ &=\frac{2\sqrt{n^2+3n+2}}{\sqrt{n+1}}-2\\ &=\frac{2\sqrt{(n+2)(n+1)}}{\sqrt{n+1}}-2\\ &=2\sqrt{n+2}-2 \end{align}

Jlamprong
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Let's prove this first: x>0, : $ 2*\sqrt{x} +\frac{1}{\sqrt{x}} > 2*\sqrt{x+1} $

This is true iif : $ (2*\sqrt{x} +\frac{1}{\sqrt{x}})^2 > (2*\sqrt{x+1})^2 $

You get : $ 2*\sqrt{x} +\frac{1}{\sqrt{x}} > 2*\sqrt{x+1} $ <=> $ 4x + \frac{1}{x} +4 > 4*(x+1) $ <=> $ \frac{1}{x} > 0 $ which is true, so the inequality is true as well :) .

Now for the induction step :

$ 2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} $

=> $2\sqrt{n+1} - 2 +\frac{1}{\sqrt{n+1}} < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}$

But we have from the first part of the answer : $2*\sqrt{n+1} +\frac{1}{\sqrt{n+1}} -2 > 2*\sqrt{n+1+1} -2 = 2*\sqrt{n+2} -2$

Hence you get: $ 2*\sqrt{n+2} -2 < 2\sqrt{n+1} - 2 +\frac{1}{\sqrt{n+1}} < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}$ Induction holds :)

mvggz
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Since the claim is true for $n=1$ it suffices to prove that at each step the increment on the left hand side is smaller than on the right hand side. But $$2\bigl(\sqrt{n+2}-\sqrt{n+1}\bigr)={2\over\sqrt{n+2}+\sqrt{n+1}}<{1\over\sqrt{n+1}}\ .$$

Christian Blatter
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