The Harmonic Series.

Murat Aygen from Turkey sent this solution to part (a).

Let us partition the terms of our series, starting from every

k = 2^{i - 1} + 1

, as follows:

\sum_{k = 1}^{n} (\frac{1}{k}) = 1 + (\frac{1}{2}) + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + \sum_{i = 4}^{n} (\frac{1}{2^{i - 1} + 1} + \dots + \frac{1}{2^{i}})

How many terms are there in each partition? We see that there are 1, 2, 4, 8, 16... terms in the successive partitions given by $2^{i} - 2^{i - 1}$ where $2^{i} - 2^{i - 1} = 2^{i - 1} (2 - 1) = 2^{i - 1}$

The smallest term in a partition is clearly the rightmost term $\frac{1}{2^{i}}$ . This smallest term multiplied by the number of terms in the partition is equal to

$\frac{1}{2^{i}} \times 2^{i - 1} = \frac{1}{2}$

which is always less than the sum of the terms of the partition. Anyway it is a positive constant! Since the number of terms in the series is as many as one wishes, we can form as many partitions as we wish whose partial sums are not less than 1/2. For reaching to 100, 200 partitions will be enough.