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Math Help - Determining series convergence or divergence

  1. #1
    Junior Member xterminal01's Avatar
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    Determining series convergence or divergence

    <br />
\sum_{n = 1}^{\infty} \frac {1}{4^n}<br />

    I tried rewriting it to 4^-n and then using the geometric series test 1/1-r
    Which makes it to -1/3 convergence. I am not sure if its correct or not...

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  2. #2
    MHF Contributor
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    This is a geometric series. Rewrite it so the index starts at zero and (1/4)^n
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  3. #3
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    I obtained a positive 1/3 when I did it.
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  4. #4
    Junior Member xterminal01's Avatar
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    Quote Originally Posted by dwsmith View Post
    I obtained a positive 1/3 when I did it.
    That probably correct can you show steps u used?
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  5. #5
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    Since you have 1/4^n, you can rewrite it as (1/4)^n. This is because 1 to any power is still 1.

    Now, because your indexing starts at 1 instead of 0, we can subtract 1 and then add 1.

    This will make the index start at 0. The adding one happens to the exponent so now we have (1/4)^{n+1}. This can be written as (1/4)^1*(1/4)^n.

    Of course, (1/4)^1=(1/4) so we just move that out side the summation.

    Thus, you have a geometric series times 1/4 which is your a in the formula a/(1-r)
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