# Series

• May 23rd 2006, 08:53 AM
Natasha1
Series

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

Hint: it might be helpful to consider $\int_{a}^{b} \surd x dx$ for suitable a and b
• May 23rd 2006, 09:57 AM
CaptainBlack
Quote:

Originally Posted by Natasha1

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

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$

So doing the integrals gives:

$
\frac{2}{3}(n)^{3/2}$

RonL
• May 23rd 2006, 12:11 PM
CaptainBlack
Quote:

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$

In explanation, this is because:

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

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

RonL
• May 23rd 2006, 12:18 PM
Natasha1
Quote:

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?
• May 23rd 2006, 01:05 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