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I am given this expression and I have to prove it using mathematical induction

${\displaystyle \forall }n\ge 1 $ , $$ \sum _{k=1}^n\left(\frac{1}{\sqrt{k}}\right)\ge 2\left(\sqrt{n+1}-1\right)$$

I have proved it is true for $n=1$, then suposed it was true until $n$ and wanted to prove it was also true until $n+1$.

I have operated a bit and arrived to this expression I want to prove: $${\frac{1}{2\sqrt{n+1}}}\ge (\sqrt{n+2}-\sqrt{n+1})$$ , and I don't know how to continue to clear that inequality to prove it. Can you help me please?? (I have just started studying this method so I'm a bit lost :/ )

Shaun
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L. Sandoval
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2 Answers2

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$$\sqrt{n + 2} - \sqrt{n + 1} = \frac{1}{\sqrt{n + 2} + \sqrt{n + 1}} \le \frac{1}{2\sqrt{n + 1}}$$ That's it.

cip999
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Since the canonical answer has already been given, let us focus on subtler things: creative telescoping produces sharper bounds. We may notice that $\frac{1}{\sqrt{k}}$ is just a bit less than $2\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right)$ (which is a telescopic term) for any $k\geq 1$, but

$$ \frac{1}{\sqrt{k}}\color{blue}{\geq} 2\left[\left(k+\tfrac{1}{2}\right)^{1/2}-\left(k-\tfrac{1}{2}\right)^{1/2}\right]-\tfrac{1}{48}\left[ \left(k+\tfrac{1}{2}\right)^{-3/2}-\left(k-\tfrac{1}{2}\right)^{-3/2}\right]$$ holds for any $k\geq 1$ just as well, and the RHS is a telescopic term. It follows that

$$ H_{n}^{(1/2)}=\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\color{blue}{\geq}2\left(n+\tfrac{1}{2}\right)^{1/2}-\tfrac{1}{48}\left(n+\tfrac{1}{2}\right)^{-3/2}-\tfrac{25}{24}\sqrt{2}. $$

Jack D'Aurizio
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