Yesterday I posed the following problem: If $S$ is the set of positive integers whose decimal expansion does not contain a 3, does<\/p>\n
\\[ \\sum_{n\\in S}\\frac{1}{n} \\]<\/p>\n
converge or diverge? My solution is after the break.<\/p>\n
<\/p>\n
The sum converges. To see this, we’ll partition $S$ into subsets $S_k$ so that each subset contains the numbers of $S$ with exactly $k$ decimal digits. For example:<\/p>\n
\\[ \\begin{align*}
\nS_1 &= \\{ 1 , 2, 4, 5, 6, 7, 8, 9 \\} \\\\
\nS_2 &= \\{ 10, 11, 12, 14, \\dotsc, 99 \\}
\n\\end{align*} \\]<\/p>\n
Note that the size of $S_k$ is $8\\cdot9^{k-1}$ since there are 8 possibilities for the first digit (anything except 0 and 3) and 9 possibilities for each of the remaining $k-1$ digits. Also note that the smallest number in $S_k$ is $10^{k-1}$, so $1\/10^{k-1}$ is an upper bound on the\u00a0reciprocal\u00a0of numbers in $S_k$.<\/p>\n
Now we split our sum using the specified partition of $S$ and use the $1\/10^{k-1}$\u00a0upper bound on each term with $k$ digits in the denominator:<\/p>\n
\\[ \\begin{align*}
\n\\sum_{n\\in S}\\frac{1}{n} &= \\sum_{k=1}^\\infty\\sum_{n\\in S_k}\\frac{1}{n} \\\\
\n&\\leq \\sum_{k=1}^\\infty\\frac{8\\cdot9^{k-1}}{10^{k-1}} \\\\
\n&= 8\\sum_{k=1}^\\infty\\Bigl(\\frac{9}{10}\\Bigr)^{k-1} \\\\
\n&= \\frac{8}{1-9\/10} \\\\
\n&= 80
\n\\end{align*} \\]<\/p>\n
Where the final summation is simplified using the summation formula for a geometric series<\/a>. By the comparison test the sum in question converges, as required.<\/p>\n