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Proove by the Complete Induction for every $n\in \mathbb{N}, n \geq 1$ $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} > \sqrt{n}$$.

I know only the basis - $\frac{1}{\sqrt{1}} \geq \sqrt{1}$.

But, I dont know how to do the change from the regular induction to the induction of Set theroy(Complete Induction).

N. F. Taussig
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NM2
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  • What is the induction of set theory as opposed to regular induction ? – Rene Schipperus May 08 '16 at 17:54
  • So you must show that if $\dfrac{1}{\sqrt{1}}+ \dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}>\sqrt{n}$ then $\dfrac{1}{\sqrt{1}}+ \dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}>\sqrt{n+1}$. That is, you must show that $\sqrt{n}+\dfrac{1}{\sqrt{n+1}}>\sqrt{n+1}$. – John Wayland Bales May 08 '16 at 17:57

1 Answers1

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I don't know about the induction of set theory but the inductive step for ordinary induction goes like this:

Assume

\begin{equation} \frac{1}{\sqrt{1}}+ \frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}>\sqrt{n} \end{equation}

\begin{equation} \sqrt{n}=\frac{n}{\sqrt{n}}>\frac{n}{\sqrt{n+1}} \end{equation} Therefore \begin{equation} \sqrt{n}>\sqrt{n+1}-\frac{1}{\sqrt{n+1}} \end{equation} so \begin{equation} \sqrt{n}+\frac{1}{\sqrt{n+1}}>\sqrt{n+1} \end{equation}

Thus \begin{equation} \frac{1}{\sqrt{1}}+ \frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}>\sqrt{n+1} \end{equation}

John Wayland Bales
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