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    Q: http://gyazo.com/5d4e59735c8f7bf2b1591119fb0e335e

    note: there is a hint in the question saying to consider the terms with even and odd indicies seperately

    I'm having troubles getting started with this question:

    Does the question want me to find a general formula for  a_n and then prove that using the root test or use  a_n = 1/2^n and prove that using the root test?

    Any help on getting started - thanks
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    (Original post by TDL)
    Q: http://gyazo.com/5d4e59735c8f7bf2b1591119fb0e335e

    note: there is a hint in the question saying to consider the terms with even and odd indicies seperately

    I'm having troubles getting started with this question:

    Does the question want me to find a general formula for  a_n and then prove that using the root test or use  a_n = 1/2^n and prove that using the root test?

    Any help on getting started - thanks
    The odd and even terms form two separate geometric series. Does that help?
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    I would think you're supposed to apply the root test directly to the given series. Note that the n^th term is K/2^n where K = 1/2 or 2; once you take nth roots the value of K isn't going to matter for large n.
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    (Original post by brianeverit)
    The odd and even terms form two separate geometric series. Does that help?
    It does but, are we allowed to just split up a_n like that? Is that not rearranging it again? I.e. can we say  \displaystyle \sum_{n=0}^\infty a_n = \sum_{n=0}^\infty \dfrac{1}{2^{2n+1}} + \sum_{n=0}^\infty \dfrac{1}{2^{2n}} ?

    (Original post by DFranklin)
    I would think you're supposed to apply the root test directly to the given series. Note that the n^th term is K/2^n where K = 1/2 or 2; once you take nth roots the value of K isn't going to matter for large n.
    Thanks that worked well for part a), however that doesn't work so well for part b: http://gyazo.com/97547384a1ac8f51553737bcc8898094

    we have  \displaystyle \sum_{n=0}^\infty \dfrac{1}{2^n} = 2 and  \displaystyle \sum_{n=0}^N \dfrac{1}{2^n} = 2 - 2^{-N}
     \displaystyle \sum_{n=0}^N a_n = \sum_{n=0}^N \dfrac{K}{2^n} = K(2-2^{-N}) and we can't necessarily conclude that is less than 2 if K can equal 2
 
 
 
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