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Math Help - seriess

  1. #16
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    Quote Originally Posted by Enita View Post
    =\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
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  2. #17
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    Re:

    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
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  3. #18
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    Quote Originally Posted by Enita View Post
    thanks!

    so is this the best way to tackle these questions?
    There is not best way. You train thy eye.
    Quote Originally Posted by Euclid
    There is no golden road to geometry.
    (Something like that).
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