Appendix - Mathematical Induction and Other Forms of Proofs - Exercises - Page A6: 8

See the proof below.

Consider the following statement: $\displaystyle\frac{1}{\sqrt{1}}+\displaystyle\frac{1}{\sqrt{2}}+\displaystyle\frac{1}{\sqrt{3}}+\ldots+\displaystyle\frac{1}{\sqrt{n}}>\sqrt{n}$ for all integers $n\geq 2$. ______________________________________________________________________________ We will use mathematical induction to show that the following equivalent statement is true. $\displaystyle\frac{1}{\sqrt{1}}+\displaystyle\frac{1}{\sqrt{2}}+\displaystyle\frac{1}{\sqrt{3}}+\ldots+\displaystyle\frac{1}{\sqrt{m+1}}>\sqrt{m+1}$ for all integers $m\geq 1$. We must show that the statement it is true for $m=1$ and that if it is true for $m=k$ then, it is true for $m=k+1$. $\textbf{1.}$ The statement is true for $m=1$ since $\displaystyle\frac{1}{\sqrt{1}}+\displaystyle\frac{1}{\sqrt{1+1}}=\displaystyle\frac{1}{\sqrt{1}}+\displaystyle\frac{1}{\sqrt{2}}=1+\displaystyle\frac{\sqrt{2}}{2}=\displaystyle\frac{2+\sqrt{2}}{2}>\displaystyle\frac{\sqrt{2}+\sqrt{2}}{2}=\sqrt{2}=\sqrt{1+1}$. $\textbf{2.}$ Suppose the statement is true for $m=k$, that is, suppose that $$\displaystyle\frac{1}{\sqrt{1}}+\displaystyle\frac{1}{\sqrt{2}}+\displaystyle\frac{1}{\sqrt{3}}+\ldots+\displaystyle\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$$. Then, adding $\displaystyle\frac{1}{\sqrt{\left(k+1\right)+1}}$ to both sides of the previous inequality gives us that $$\displaystyle\frac{1}{\sqrt{1}}+\displaystyle\frac{1}{\sqrt{2}}+\displaystyle\frac{1}{\sqrt{3}}+\ldots+\displaystyle\frac{1}{\sqrt{k+1}}+\displaystyle\frac{1}{\sqrt{\left(k+1\right)+1}}>\sqrt{k+1}+\displaystyle\frac{1}{\sqrt{\left(k+1\right)+1}}.$$ Observe that $$\sqrt{k+1}+\displaystyle\frac{1}{\sqrt{\left(k+1\right)+1}}=\sqrt{k+1}+\displaystyle\frac{1}{\sqrt{k+2}}=$$ $$\displaystyle\frac{\sqrt{k+1}\sqrt{k+2}+1}{\sqrt{k+2}}>\displaystyle\frac{\sqrt{k+1}\sqrt{k+1}+1}{\sqrt{k+2}}=\displaystyle\frac{k+1+1}{\sqrt{k+2}}=$$ $$\displaystyle\frac{k+2}{\sqrt{k+2}}=\sqrt{k+2}=\sqrt{\left(k+1\right)+1}.$$ which gives us that $$\displaystyle\frac{1}{\sqrt{1}}+\displaystyle\frac{1}{\sqrt{2}}+\displaystyle\frac{1}{\sqrt{3}}+\ldots+\displaystyle\frac{1}{\sqrt{k+1}}+\displaystyle\frac{1}{\sqrt{\left(k+1\right)+1}}>\sqrt{\left(k+1\right)+1}.$$ Thus, the statement holds for $m=k+1$ whenever it is true for $m=k$. Hence, the mathematical induction principle guarantees that the statemente is true for every positive integer $m$. ____________________________________________________________ Finally, given any positive integer $n\geq 2$ we have that $m=n-1\geq 1$ and $n=\left(n-1\right)+1=m+1$. Thus it follows from the statement we have just proved that $$\displaystyle\frac{1}{\sqrt{1}}+\displaystyle\frac{1}{\sqrt{2}}+\displaystyle\frac{1}{\sqrt{3}}+\ldots+\displaystyle\frac{1}{\sqrt{n}}>\sqrt{n}$$ is true for all integers $n\geq 2$.