Problem: Find all positive integers \(a,\ n\) such that \[\frac{1}{a} + \frac{1}{a+1}+ \cdots \frac{1}{a+n}\] is an integer.

Solution: There does not exist such a pair \((a, n)\).

Let \(i\) be an integer with \(0 \leq i \leq n\).

Case 1: Suppose that no \(a + i\) is of the form \(2^\alpha\). Then, there exists an integer \(k\) such that \[ 2^k \lt a + 1,\ \cdots,\ a + n \lt 2^{k+1} \] and \[\frac{1}{a} + \cdots \frac{1}{a+n} \lt \frac{1}{2^k}+\cdots +\frac{1}{2^{k}} \lt \frac{1}{2^{k}}(2^{k+1} - 2^k) = 1.\] Therefore, \(\frac{1}{a} + \frac{1}{a+1}+ \cdots \frac{1}{a+n}\) can not be an integer.

Case 2: Suppose that one of \(a + i\) is of the form \(2^\alpha\). Let \(m = \max\{ \alpha : 2^\alpha \parallel (a + i), 0 \leq i \leq n\}\). Note that if \(2^m \parallel (a + i)\), then \(2^m = a + i\), and such \(i\) is unique, namely \(i_m\). (If not, then there exists an \(i'\) such that \(2^{m+1} \parallel (a + i')\), which contradicts the definition of \(m\).) Let \(d\) be \(d = [a,\ \cdots,\ a+n]\), that is, the least common multiple of \(a,\ \cdots,\ a+n\). Since \(d\) is even, if \[\frac{1}{a} + \cdots \frac{1}{a+n} = \frac{d/a + \cdots d/(a+n)}{d}\] is an integer, then the numerator \(d/a + \cdots d/(a+n)\) must also be an even number. However, this is impossible, since \({d}/({a+i})\) is even for all \(i \neq i_m\), and \(d/(a + i_m)\) is the only odd term in the sum.

Problem: Show that \[\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\frac{1}{k2^{n}+1} =1.\]

Solution1: Let \(a_k\) be \(a_k = \frac{1}{k}\sum_{n=0}^{\infty}{\frac{1}{k2^{n}+1}}\). Note that the summand \(|(-1)^{k-1}a_k| = O(1/k^2)\). Therefore, the given series may be arbitrarily rearranged. That is, \[\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}{\frac{1}{k2^{n}+1}}} = \sum_{k=1}^{\infty}{(a_k - 2a_{2k})}\] \[=\sum_{k=1}^{\infty} \left[ \frac{1}{k}\sum_{n=0}^{\infty}{\frac{1}{k2^{n}+1}} -2\times\frac{1}{2k}\sum_{n=0}^{\infty}{\frac{1}{k2^{n+1}+1}} \right] \] \[=\sum_{k=1}^{\infty}\frac{1}{k} \left[ \sum_{n=0}^{\infty}\frac{1}{k2^{n}+1} - \sum_{n=1}^{\infty}\frac{1}{k2^{n}+1} \right]\] \[=\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=1.\]

Solution2: \[\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\frac{1}{k2^{n}+1} = \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\int_{0}^{1}x^{k2^{n}}dx\] \[=\int_{0}^{1}\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}x^{k2^{n}}dx=\int_{0}^{1}\sum_{n=0}^{\infty}\log (1+x^{2^n})dx\] \[=\int_{0}^{1}\log \prod_{n=0}^{\infty}(1+x^{2^n})dx=\int_{0}^{1}\log\frac{1}{1-x}dx = 1\]