A New Year’s Puzzle

Happy New Year!

Now that you’ve caught up on your sleeping and filled yourself with too much food, you probably have a hankerin’ for a math problem.  Here is one we particularly like because it only uses Calc III sequence and series stuff:

Everybody knows that the harmonic series

$\sum_{n=1}^{\infty} \frac{1}{n}$

diverges (since it’s partial sums grow without bound, it can’t converge to any fixed number).  One amusing consequence of this is the fact that you can build a bridge from New York to London by simply stacking playing cards on top of each other.

Now let’s consider a closely related series.  Pick your favorite digit between 2 and 9.  We’ll pick 2.  Make a new series by taking the harmonic series and deleting all terms which have your fixed digit in it one or more times.  So we would delete 1/2, 1/12, 1/20, 1/21, 1/22, …., 1/8591721, etc.   Of course, the majority of the terms of the harmonic series will remain undeleted.   This new series is the one we want to think about.

Question:  Does this new series converge?

Question:  Does whether or not it converge depend on which digit you delete?

Question:  If it converges, say to the number S, you know that it can be very difficult to compute what S is for most series.  Instead, can you at least give a number which you know is larger than S?

Solutions are welcome in the comments.  Hint:  The answer to these questions only require Calc III mathematics. We’ll post the solution in a few days or, if you’re impatient, you can google the solution without too much difficulty.

Below is an interesting related question:

On the other hand, we all know that the series

$\sum_{n=2}^{\infty} \frac{1}{n^2} = \sum_{n=2}^{\infty} \left( \frac{1}{n} \right)^2$

converges since it’s a p-series with $p=2$.  In this case we actually know (thanks to Euler) the series converges to $S=\pi^2/6 - 1$.   Here is an amusing question:  Notice that $S \leq (5/6)^2$.   If you think of the above sum as the areas of squares of side length 1/2, 1/3, 1/4, etc., can you tile a square of side length 5/6 with one square of each of these sizes without overlapping the squares?

The answer is yes.  This pictures shows how to do it (in the picture n stands for the square of side length 1/n):

Tiling a Square thanks to Erich Friedman

You can find more details here.