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# Prove each statement by mathematical induction.$\sqrt { n } < \frac { 1 } { \sqrt { 1 } } + \frac { 1 } { \sqrt { 2 } } + \cdots + \frac { 1 } { \sqrt { n } }$, for all integers$n \geq 2$.

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To proof: $\sqrt{n}<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{n}}$, for all integers $n\geq 2$

$\textbf{PROOF BY INDUCTION}$

Let $P(n)$ be " $\sqrt{n}<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{n}}$"

$\textbf{Basis step}$ $n=2$

\begin{align*} \sqrt{n}&=\sqrt{2}\approx \color{#c34632}1.4 \\ \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{n}}&=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}=1+\dfrac{\sqrt{2}}{2}\approx \color{#c34632}1.7 \end{align*}

Thus $P(2)$ is true, since $1.4<1.7$.

$\textbf{Inductive step}$ Let $P(k)$ be true, thus $\sqrt{k}<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{k}}$

We need to prove that $P(k+1)$ is true.

\begin{align*} \sqrt{k+1}&=\sqrt{k+1}+\sqrt{k}-\sqrt{k} \\ &=\sqrt{k}+(\sqrt{k+1}-\sqrt{k}) \\ &<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{k}}+(\sqrt{k+1}-\sqrt{k})&\color{#4257b2}\text{Since P(k) is true} \end{align*}

We need to show that $\sqrt{k+1}<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+....+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}$, by the above derivation it then suffices to show $\sqrt{k+1}-\sqrt{k}<\dfrac{1}{\sqrt{k+1}}$ (as the strict inequality was already used in the above derivation).

$\sqrt{k+1}-\sqrt{k}< \dfrac{1}{\sqrt{k+1}}$ is equivalent with $\sqrt{k+1}(\sqrt{k+1}-\sqrt{k})<1$ (by multiplying each side by $\sqrt{k+1}$, or using the distributive property $k+1-\sqrt{k+1}\sqrt{k}<1$ (which we will still have to proof).

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