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?

Question

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?

freckles · Answer

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

anonymous · Answer

Yeah the digits in the ones place wouldn't be very interesting, I'm pretty sure they're all $0$ after a certain point.

freckles · Answer

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...

freckles · Answer

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

anonymous · Answer

Yep that's right.

freckles · Answer

didn't do the reciprocals

freckles · Answer

omg and I don't know how to count

freckles · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

i must be missing something 
seems silly to me

freckles · Answer

that does sound right to me

anonymous · Answer

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