0
$\begingroup$

I want to prove, using mathematical induction, the following proposition:

$$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\:\ge \sqrt{n}, \forall n \geq 1 \in \mathbb{N}$$

My thesis is:

$$\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}, \forall n \geq 1 \in \mathbb{N}$$

I've proved the inequality for $n=1$, but after that I'm not being able to do the rest :/

Thank you for the help!

$\endgroup$
18
  • 1
    $\begingroup$ false............. The sum is larger than $\sqrt n,$ it is approximately $2 \sqrt n$ minus a modest constant $\endgroup$
    – Will Jagy
    Sep 30 '17 at 18:03
  • $\begingroup$ @WillJagy I'm sorry, it's now corrected! $\endgroup$ Sep 30 '17 at 18:05
  • $\begingroup$ What do you think you should do? I don't see any of your writing or thinking, so I don't know what you're stuck on. Have you even tried to show the base case ($n=2$)? $\endgroup$ Sep 30 '17 at 18:10
  • $\begingroup$ @EricTowers Yes, i've done that, and i'm stuck after that, I've study mathematical induction but i'm not understand how can i apply that for this example $\endgroup$ Sep 30 '17 at 18:12
  • 1
    $\begingroup$ Ok this is a start. Completely write down the inequality for $n+1$ and review your thesis. $\endgroup$ Sep 30 '17 at 18:49
1
$\begingroup$

For base cases $n=1$ we get that $1 \geq \sqrt{1}$

For $n=2$ we get that $1+\frac{1}{\sqrt{2}} \geq \sqrt{2}$.

Assume that $1+\frac{1}{\sqrt{2}} +\cdots +\frac{1}{\sqrt{n}} \geq \sqrt{n}$.

And we want to prove that $1+\frac{1}{\sqrt{2}} +\cdots +\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$

From the assumption step we know that $1+\frac{1}{\sqrt{2}} +\cdots +\frac{1}{\sqrt{n}} \geq \sqrt{n}$

Substitute instead of $1+\frac{1}{\sqrt{2}} +\cdots +\frac{1}{\sqrt{n}}$ the $\sqrt{n}$.

To get $\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$ (This what you need to prove).

$\endgroup$
1
  • $\begingroup$ thank you very much for the help! And how can I prove that last line? I've done the sum but it's giving me something really different that what we need to prove :O $\endgroup$ Sep 30 '17 at 19:14
1
$\begingroup$

The Inductive Step

Since $$ \begin{align} \sqrt{n}-\sqrt{n-1} &=\frac1{\sqrt{n}+\sqrt{n-1}}\\ &\le\frac1{\sqrt{n}}\tag1 \end{align} $$ we have $$ \begin{align} \sum\limits_{k=1}^n\frac1{\sqrt{k}} &=\color{#C00}{\sum\limits_{k=1}^{n-1}\frac1{\sqrt{k}}}+\frac1{\sqrt{n}}\tag2\\ &\ge\color{#C00}{\sqrt{n-1}}+\frac1{\sqrt{n}}\tag3\\[9pt] &\ge\sqrt{n}\tag4 \end{align} $$ Explanation:
$(2)$: expand sum
$(3)$: inductive hypothesis
$(4)$: apply $(1)$

$\endgroup$
0
$\begingroup$

$$\sqrt n\;+\frac {1}{\sqrt {n+1}}\geq \sqrt {n+1}\iff \frac {1}{\sqrt {n+1}}\;\geq \sqrt {n+1}-\sqrt n=$$ $$=(\sqrt {n+1}-\sqrt n)\cdot \frac {\sqrt {n+1}+\sqrt n}{\sqrt {n+1}+\sqrt n}=\frac {1}{\sqrt {n+1}+\sqrt n}\iff$$ $$\iff \frac {1}{\sqrt {n+1}}\geq \frac {1}{\sqrt {n+1}+\sqrt n}\;.$$

$\endgroup$
0
$\begingroup$

For $n \ge1 $, we have $$n \ge n-1 + \frac1{n+1}$$ $$\iff n \ge n+1-2 + \frac1{n+1}$$ $$\Rightarrow n \ge (\sqrt{n+1} - \frac1{\sqrt{n+1}})^2$$ $$\Rightarrow \sqrt{n} \ge \sqrt{n+1} - \frac1{\sqrt{n+1}}$$

Now $\sqrt{1} \ge 1$. Assume $\sum_{r=1}^n \frac1{\sqrt{r}} \ge \sqrt{n}$.

Then $$\sum_{r=1}^n \frac1{\sqrt{r}} \ge \sqrt{n} \ge \sqrt{n+1} - \frac1{\sqrt{n+1}}$$

$$\Rightarrow \sum_{r=1}^{n+1} \frac1{\sqrt{r}} \ge \sqrt{n+1}$$

Hence , by induction , $\forall n \in \mathbb{N}, \sum_{r=0}^n \frac1{\sqrt{r}} \ge \sqrt{n}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.