What can be said about this sequence:

$\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

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- May 23rd 2006, 07:53 AMNatasha1Series
What can be said about this sequence:

$\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 AMCaptainBlackQuote:

Originally Posted by**Natasha1**

$\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 AMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

$\displaystyle

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

$

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

RonL - May 23rd 2006, 11:18 AMNatasha1Quote:

Originally Posted by**CaptainBlack**

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 PMCaptainBlackQuote:

Originally Posted by**Natasha1**

RonL