complex sum of reciprocal decimal expansions terminating in odd digits Let bn be the sequence of all positive integers such that the decimal expression for 
1/bn
terminates in an odd digit:
1, 2, 4, 8, 10, . . .
(For instance, 3 is not included because 1/
3 = 0.33 . . . does not terminate, 4 is included
because 1/
4 = 0.25 terminates in 5, which is odd; 5 is not included because 1/
5 = 0.2
terminates in 2, which is even.)
Find sum to infinity 1/bn
 A: It's well known and easy to prove that the only $\frac 1{b_i}$ that terminate are $b_i$ of the form $b_i = 2^k 5^m$.  
(Because the only divisors of powers of $10$ are $2$ and $5$ so any other prime will never divide into $10^N$ evenly.)
If $b_i = 2^k 5^m$ then either:
1) $k > m$ and $b_i = 2^{k-m}*10^m$ and this will terminate in $5$.
2) $k = m$ and $b_i = 10^m$ and this will terminate in $1$.
or 3) $k < m$ and $b_i = 5^{m-k}*10^k$ and this will terminate with a power of $2$.
3) are not an acceptable option but 1) and 2) are. 
So we are dealing with numbers of the form $b_i = 2^j*10^k$ (where $j$ and $k$ may be $0$).
So you sum will be
$\sum_{j=0; k= 0}^{\infty, \infty} \frac 1{2^j}*\frac 1{10^k}$.... IF this converges we can rearrange these to
$\sum_{k=0}^{\infty}(\frac 1{10^k}\sum_{j=0}^{\infty}\frac 1{2^j})$
Now you should know that $\sum_{j=0}^{\infty}\frac 1{2^j} = 1 + \frac 12 + \frac 14 + .... =2$.
So ..... you get $\sum_{k=0}^{\infty} \frac 1{10^k}*2 = 2.22222222....... = 2 \frac 29$
