Series

**Natasha1** · May 23rd 2006, 07:53 AM

What can be said about this sequence:

$1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n$

Hint: it might be helpful to consider $\int_{a}^{b} \surd x dx$ for suitable a and b

**CaptainBlack** · May 23rd 2006, 08:57 AM

Originally Posted by Natasha1

What can be said about this sequence:

$1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n$

Hint: it might be helpful to consider $\int_{a}^{b} \surd x dx$ for suitable a and b

Let:

$ S_n= 1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n $

Then as $\{\sqrt{r}, r=1,2, \dots,\ n\}$ is an increasing sequence:

$ \int_{x=1}^{n+1} \sqrt{x-1} dx<S_n<\int_{x=1}^{n+1} \sqrt{x} dx $

So doing the integrals gives:

$ \frac{2}{3}(n)^{3/2}<S_n<\frac{2}{3} ((n+1)^{3/2}-1) $

RonL

**CaptainBlack** · May 23rd 2006, 11:11 AM

Originally Posted by CaptainBlack

Let:

$ S_n= 1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n $

Then as $\{\sqrt{r}, r=1,2, \dots,\ n\}$ is an increasing sequence:

$ \int_{x=1}^{n+1} \sqrt{x-1} dx<S_n<\int_{x=1}^{n+1} \sqrt{x} dx $

In explanation, this is because:

$ \sqrt{x-1}<\sqrt{n}<\sqrt{x} $

for $x \in (n,n+1)$

RonL

**Natasha1** · May 23rd 2006, 11:18 AM

Originally Posted by CaptainBlack

In explanation, this is because:

$ \sqrt{x-1}<\sqrt{n}<\sqrt{x} $

for $x \in (n,n+1)$

RonL

RonL,

At the bottom of your integrals you put x=1, should it not just be 1? Could I just put 1 or not?

**CaptainBlack** · May 23rd 2006, 12:05 PM

Originally Posted by Natasha1

RonL,

At the bottom of your integrals you put x=1, should it not just be 1? Could I just put 1 or not?

Yes you can put 1

RonL

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