$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} \leq 2\sqrt{n}$ by mathematical induction

Question

I need the following by the principle of Mathematical induction:

\begin{align*} 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} \leq 2\sqrt{n} \end{align*} I can easily find out the base case for n=1 is true.

However, I can't find the proof in the induction step.

I assume: \begin{align*} 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \leq 2\sqrt{n} \end{align*}

To proof: \begin{align*} 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} \leq 2\sqrt{n+1} \end{align*} Can someone help me find a way to proof the inductive step too?

Thank you!

Answer 1

Based on the induction assumption one has \begin{align*} 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} \leq 2\sqrt{n} + \frac{1}{\sqrt{n+1}} \leq 2\sqrt{n+1} \end{align*}

The last step takes place because \begin{align*} 2\sqrt{n+1} - 2\sqrt{n} - \frac{1}{\sqrt{n+1}} & = 2(\sqrt{n+1} - \sqrt{n}) - \frac{1}{\sqrt{n+1}} = \frac{2}{\sqrt{n+1}+\sqrt{n}} - \frac{1}{\sqrt{n+1}}\\\\ & = \frac{2\sqrt{n+1} - \sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}+\sqrt{n})\sqrt{n+1}} = \frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}+\sqrt{n})\sqrt{n+1}} > 0 \end{align*}

Answer 2

Hint: Like any induction proof, use the induction hypothesis. You have an assumption. That assumption lets you simplify away almost all of the left-hand side, at the cost of making it a little bit bigger. Hopefully it is still smaller than $2\sqrt{n+1}$, though.

Answer 3

The increment from the $n$ sum to the $n+1$ sum is

$\frac{1}{\sqrt{n+1}}=\frac{2}{\color{blue}{\sqrt{n+1}}+\sqrt{n+1}}<\frac{2}{\color{blue}{\sqrt{n}}+\sqrt{n+1}}$

and then

$\frac{2}{\sqrt{n}+\sqrt{n+1}}=2(\sqrt{n+1}-\sqrt{n})$

from the difference of squares factorization. Can you see how to do the induction now?