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anonymous

  • one year ago

Consider the sequence \(a_n\) whose \(n\)th term is given by the reciprocal of the first digit of \(n!\). For example, \(a_1=1\) since the first digit of \(1!=1\) is \(1\); \(a_6=\dfrac{1}{7}\) since \(6!=720\), and so on. Take another sequence, \(b_n\), a subsequence of \(a_n\) that doesn't contain \(1\). (This measure is taken just in case there are infinitely many terms of \(a_n\) that are \(1\).) Does \(\displaystyle\sum_{n=1}^\infty b_n\) converge?

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  1. freckles
    • one year ago
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    darn it always get confused about 1st and last we are definitely not looking at the units digit

  2. anonymous
    • one year ago
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    Yeah the digits in the ones place wouldn't be very interesting, I'm pretty sure they're all \(0\) after a certain point.

  3. freckles
    • one year ago
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    yep that is how I figured out it definitely wasn't the units digit and then I also looked at your example after noticing the factor 2*5 for a_5,a_6 and so on...

  4. freckles
    • one year ago
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    so say the sequence you are talking about is \[a_n=1,2,6,2,1,7,5,4,3,3,3,4,6 \text{ then } b_n \text{ would be } 2,6,2,7,5,4,3,3,3,4,6\] those are the first 9 terms of the an sequence

  5. anonymous
    • one year ago
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    Yep that's right.

  6. freckles
    • one year ago
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    didn't do the reciprocals

  7. freckles
    • one year ago
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    omg and I don't know how to count

  8. freckles
    • one year ago
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    ok I'm just going to be quiet until I can point out something awesome if that ever comes around problem might be too hard for old freckles

  9. anonymous
    • one year ago
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    Ugh, accidentally closed the tab mid-typing...

  10. anonymous
    • one year ago
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    i am either very confused, or there is no way for this to converge probably the former

  11. anonymous
    • one year ago
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    if i am reading it correcttly (probably not) it is bounded below by \(\sum \frac{1}{9}\)

  12. anonymous
    • one year ago
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    i must be missing something seems silly to me

  13. freckles
    • one year ago
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    that does sound right to me

  14. anonymous
    • one year ago
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    No, you're absolutely right @satellite73. I didn't realize this question was far more trivial than it had seemed at first...

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