seriess

**topsquark** · January 27th 2007, 05:29 AM

Originally Posted by Enita

$=\left| \frac{1+2n^2+4n+2}{1+n^2+2n+1} \cdot \frac{1+n^2}{1+2n^2} \right|=\left| \frac{4n+2}{2n+1}\right|$

divde by n
$=\left| \frac{4+2}{2+1}\right|<br />$

Series equals 6/3 which is > 1 so diverges?

Careful! You can't cancel the n's here since n is not a factor of either the numerator, nor the denominator.

What you can do is this. Divide both the numerator and denominator by n:

$\left | \frac{4n+2}{2n+1} \right | = \left | \frac{4 + \frac{2}{n}}{2 + \frac{1}{n}} \right |$

Now, since n is large this is approximately:

$\left | \frac{4n+2}{2n+1} \right | = \left | \frac{4 + \frac{2}{n}}{2 + \frac{1}{n}} \right | \approx \left | \frac{4}{2} \right | = 2$

Which shows that the series diverges.

-Dan

**Enita** · January 27th 2007, 10:16 AM

thanks!

so is this the best way to tackle these questions? (as it is an approximation) or is there an alternate method too?

thanks again

**ThePerfectHacker** · January 27th 2007, 02:12 PM

Originally Posted by Enita

thanks!

so is this the best way to tackle these questions?

There is not best way. You train thy eye.

Originally Posted by Euclid

There is no golden road to geometry.

(Something like that).

Thread: seriess

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