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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?
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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freckles
 one year ago
Best ResponseYou've already chosen the best response.0darn it always get confused about 1st and last we are definitely not looking at the units digit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah the digits in the ones place wouldn't be very interesting, I'm pretty sure they're all \(0\) after a certain point.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0yep 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...

freckles
 one year ago
Best ResponseYou've already chosen the best response.0so 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.0didn't do the reciprocals

freckles
 one year ago
Best ResponseYou've already chosen the best response.0omg and I don't know how to count

freckles
 one year ago
Best ResponseYou've already chosen the best response.0ok 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ugh, accidentally closed the tab midtyping...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am either very confused, or there is no way for this to converge probably the former

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if i am reading it correcttly (probably not) it is bounded below by \(\sum \frac{1}{9}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i must be missing something seems silly to me

freckles
 one year ago
Best ResponseYou've already chosen the best response.0that does sound right to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, 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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