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Infinite series problem

  1. Nov 19, 2006 #1
    Hello i have the infinite series

    7^(K+1)/2^(3k-1)

    How do i find what it converges to if it does converge.

    Limit comparison does me no good. I am thinking integral and ratio test.

    root test does me no good either.
     
  2. jcsd
  3. Nov 19, 2006 #2
    Think geometric series, the ratio test also works nicely and is probably easier than trying to make this look more like a geometric series. The root test should work as well, but I think it would be a little tricky.
     
    Last edited: Nov 19, 2006
  4. Nov 19, 2006 #3

    Office_Shredder

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    A simpler way, see if you can re-write it as a series in terms of (7/8)k
     
  5. Nov 22, 2006 #4
    Thanks for the replies, I applied the ratio test.

    7^K+1+1 2^3k-1 7^k (x) 7^2 2^3k (x) 2^-1
    --------- x ---------- = ----------------- x --------------
    2^3k-1+1 7^k+1 2^3k 7^k (x) 7^1

    -----------
    7^k+1
    ------
    2^3k-1


    So everything but the 7^2 which is 49 and 2^-1 / 7 which is 24.5 / 7 , which gives me 3.5, however i think this is wrong.
     
  6. Nov 22, 2006 #5

    HallsofIvy

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    You say "thanks for the replies" but simply ignore them?

    Your calculation is completely wrong:
    7^K+1+1 2^3k-1 7^k (x) 7^2 2^3k (x) 2^-1
    --------- x ---------- = ----------------- x --------------
    2^3k-1+1 7^k+1 2^3k 7^k (x) 7^1
    [tex]\frac{7^{k+1+1}}{2^{3k-1+1}}[/tex]
    is wrong. You are adding 1 to k, not to the exponent. It should be:
    [tex]\frac{7^{(k+1)+1}}{2^{3(k+1)-1}}= \frac{7^{k+2}}{2^{3k+2}}[/tex].

    In any case, it is far simpler to do as both d_leet and Office Shredder suggested: write this as a geometric sequence with common ratio 7/8. That way, it is not only obvious that the sequence converges but easy to see what it converges to!
     
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