# Series

• May 23rd 2006, 07:53 AM
Natasha1
Series

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

Hint: it might be helpful to consider $\displaystyle \int_{a}^{b} \surd x dx$ for suitable a and b
• May 23rd 2006, 08:57 AM
CaptainBlack
Quote:

Originally Posted by Natasha1

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

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

Let:

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

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

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

So doing the integrals gives:

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

RonL
• May 23rd 2006, 11:11 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Let:

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

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

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

In explanation, this is because:

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

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

RonL
• May 23rd 2006, 11:18 AM
Natasha1
Quote:

Originally Posted by CaptainBlack
In explanation, this is because:

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

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

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?
• May 23rd 2006, 12:05 PM
CaptainBlack
Quote:

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