## 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?

1. freckles

darn it always get confused about 1st and last we are definitely not looking at the units digit

2. anonymous

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

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

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

Yep that's right.

6. freckles

didn't do the reciprocals

7. freckles

omg and I don't know how to count

8. freckles

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

Ugh, accidentally closed the tab mid-typing...

10. anonymous

i am either very confused, or there is no way for this to converge probably the former

11. anonymous

if i am reading it correcttly (probably not) it is bounded below by $$\sum \frac{1}{9}$$

12. anonymous

i must be missing something seems silly to me

13. freckles

that does sound right to me

14. anonymous

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