Proof by Induction : Inequalities

**firebio** · February 15th 2010, 09:10 AM

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

K+1 Step

$ 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}} $

$ 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.
Thanks in advance

**Nyrox** · February 15th 2010, 11:43 AM

You left out a term in the last line. It should read

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

What you need to show now is that:

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

or equivalently, that

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

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

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

I'll let you conclude

**firebio** · February 15th 2010, 12:06 PM

Thanks for help
I think i get it

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

Then

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

That means

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

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

So $K+1 \in\mathbb N$

**HallsofIvy** · February 16th 2010, 03:35 AM

"so $K+1\in N$ "? 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".

**Archie Meade** · February 16th 2010, 04:59 AM

Originally Posted by firebio

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

K+1 Step

$ 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}} $

$ 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.
Thanks in advance

There is no need to bring in $\le$

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

Therefore, the following should be true

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

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

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

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

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

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

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

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.

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