How to prove an inequality through induction

Question

I am taking real analysis this semester and am confused on how to prove this inequality. It is

$\sqrt{2k+1} - 1 < 1 + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{2k-1}} \leq \sqrt{2k-1} $

I was thinking about trying to break the inequalities apart, and do two at at time in succession, proving the RHS and middle inequality. However, I'm not sure its even possible to prove through induction that an equation holds an inequality with a sum. So I am not sure how to approach this. Any advice would be greatly appreciated

Answer 1

@WillJagy's comment gives a nice way to do it, but since you seem to require induction, let's have it. I'll prove the right hand side inequality and leave the left hand side to do, it should be just as straightforward.

• $k=1$ : $ 1 \leq 1$ so the base case is satisfied.
• Assume that $ 1+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{2k-1}}\leq \sqrt{2k-1}$ holds for some $k>1$. Now using this we get \begin{align*} 1+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{2k-1}}+\frac{1}{\sqrt{2k+1}}\leq \sqrt{2k-1}+\frac{1}{\sqrt{2k+1}}\end{align*}

We need to prove that $\sqrt{2k-1}+\frac{1}{\sqrt{2k+1}}\leq \sqrt{2k+1}$, a quick calculation shows this is true for $k\geq \frac{1}{2}$ so true for any positive integer, and we're done.