# Proof by Induction : Inequalities

• Feb 15th 2010, 08:10 AM
firebio
Proof by Induction : Inequalities
Proof by induction that
$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{n}}< 2\sqrt{n}$

K+1 Step

$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{k}}+\frac{1}{\sqrt{k+1}}< 2\sqrt{k}+\frac{1}{\sqrt{k+1}}$

$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{k}}+\frac{1}{\sqrt{k+1}}<\frac{2\sqrt{k} \sqrt{k+1}}{\sqrt{k+1}}$

Then im stuck here....

Any help would be appreciated.
• Feb 15th 2010, 10:43 AM
Nyrox
You left out a term in the last line. It should read

$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{k}}+\frac{1}{\sqrt{k+1}}<\frac{2\sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}}$

What you need to show now is that:

$\displaystyle \frac{2\sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}}\leq 2\sqrt{k+1}$,

or equivalently, that

$\displaystyle 2\sqrt{k} \sqrt{k+1}\leq 2k+1$.

Squaring both sides (both sides are positive, as $\displaystyle k\in\mathbb N$) we get

$\displaystyle 4k(k+1)\leq (2k+1)^2$.

I'll let you conclude :)
• Feb 15th 2010, 11:06 AM
firebio
Thanks for help
I think i get it

so
$\displaystyle 4k(k+1)\leq (2k+1)^2$

Then

$\displaystyle 4k^2+4k \leq 4k^2+4k+1$

That means

$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{k}}+\frac{1}{\sqrt{k+1}}<\frac{2\sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}} \leq$
$\displaystyle 2\sqrt{k+1}$

Finally
$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{k}}+\frac{1}{\sqrt{k+1}}<$
$\displaystyle 2\sqrt{k+1}$

So $\displaystyle K+1 \in\mathbb N$
• Feb 16th 2010, 02:35 AM
HallsofIvy
"so $\displaystyle K+1\in N$"? There was no "K" in what you did before. If you mean "k", that is not what you were trying to prove and there is no need to say it- since "k" was in N, certainly k+1 is.

Your last line should be "so the statement is true for all n".
• Feb 16th 2010, 03:59 AM
Quote:

Originally Posted by firebio
Proof by induction that
$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{n}}< 2\sqrt{n}$

K+1 Step

$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{k}}+\frac{1}{\sqrt{k+1}}< 2\sqrt{k}+\frac{1}{\sqrt{k+1}}$

$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{k}}+\frac{1}{\sqrt{k+1}}<\frac{2\sqrt{k} \sqrt{k+1}}{\sqrt{k+1}}$

Then im stuck here....

Any help would be appreciated.

There is no need to bring in $\displaystyle \le$

$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{k}}< 2\sqrt{k}$

Therefore, the following should be true

$\displaystyle 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1} {\sqrt{k}}+\frac{1}{\sqrt{k+1}}< 2\sqrt{k+1}$

We try to prove this using the first statement, since that will link every pair of adjacent natural numbers in a chain.

$\displaystyle 2\sqrt{k}+\frac{1}{\sqrt{k+1}}=\frac{2\sqrt{k}\sqr t{k+1}}{\sqrt{k+1}}+\frac{1}{\sqrt{k+1}}=\frac{2\s qrt{k^2+k}+1}{\sqrt{k+1}}$

Is $\displaystyle \frac{2\sqrt{k^2+k}+1}{\sqrt{k+1}}<2\sqrt{k+1}$ ?

$\displaystyle 2\sqrt{k^2+k}+1<2(k+1)$ ?

$\displaystyle 2\sqrt{k^2+k}<2k+1$ ?

$\displaystyle \sqrt{4k^2+4k}<\sqrt{4k^2+4k+1}$ ?

Yes.

The adjacent-term chain is established.
Finally prove the inequality is true for an initial value of n.
Then it's true for that n and all following n.