Following are some interesting problems from The American Mathematical Monthly and Real infinite series(2006).

Problem 1

AMM 11665

Let $a, b$ be positive real numbers. Prove that $a^b + b^a > 1$.

Proof:
If $a > 1$ (or $b > 1$), than $a^b > 1$ (or $b^a > 1$ correspondingly). Otherwise, we assume that $a, b < 1$. Following from Bernoulli’s inequality, we have

$$ a^{1-b} = (1 + (a - 1))^{1-b} < 1 + (1-b)(a-1) = a + b - ab. $$

Hence $a^b > \frac{a}{a + b - ab}$, and

$$ a^b + b^a > \frac{a}{a + b - ab} + \frac{b}{a + b - ab} > 1. $$


Problem 2

Gem 83 in Real infinite series

For each integer $m > 0$, let $L(m)$ be the number of digits in the decimal representation of $m$ that are at least 5. Then

$$ \sum_{t=0}^{\infty} \frac{L(2^t)}{2^t} = \frac{2}{9}. $$

Proof:
Let $L_k(m)$ be 1 if the $k$-th digit of $m$ is at least 5 and 0 otherwise. Then by definition

$$ L(m) = \sum_{k=1}^{\infty} L_k(m). $$

Note that

$$ L_k(m) = \left\lfloor \frac{m}{10^k} + \frac{1}{2} \right\rfloor - \left\lfloor \frac{m}{10^k} \right\rfloor = \left\lfloor \frac{2m}{10^k} \right\rfloor - 2 \left\lfloor \frac{m}{10^k} \right\rfloor, $$

where we have used $\lfloor x + 0.5 \rfloor = \lfloor 2x \rfloor - \lfloor x \rfloor$.

Therefore, we have

$$\begin{align} \sum_{t=0}^{\infty} \frac{L(2^t)}{2^t} &= \sum_{t=0}^{\infty} \sum_{k=1}^{\infty} \frac{L_k(2^t)}{2^t} = \sum_{k=1}^{\infty} \sum_{t=0}^{\infty} \left( \frac{1}{2^t} \left\lfloor \frac{2^{t+1}}{10^k} \right\rfloor - \frac{1}{2^{t-1}} \left\lfloor \frac{2^{t}}{10^k} \right\rfloor \right) \\ &= \sum_{k=1}^{\infty} \left( \lim_{n \rightarrow \infty} \frac{1}{2^n} \left\lfloor \frac{2^{n+1}}{10^k} \right\rfloor \right) = \sum_{k=1}^{\infty} \frac{2}{10^k} = \frac{2}{9}. \end{align}$$


Problem 3

Gem 86 in Real infinite series

Let $a_1, a_2, a_3, \cdots$ be integers greater than 1, and $r$ be an integer not less than 2. Prove that

$$ x = \frac{1}{a_1} + \frac{1}{a_1^r a_2} + \frac{1}{a_1^r a_2^r a_3} + \cdots $$

is a rational number if and only if the sequence $\{ a_n \}$ is eventually periodic, i.e.

$$ \exists m, k, ~ \forall i > m, ~ a_{i + k} = a_i. $$

Proof:
At first, it is easy to check that if the sequence $\{ a_n \}$ is eventually periodic, then $x$ is a rational number.
Set

$$ z_i = \frac{1}{a_i} + \frac{1}{a_i^r a_{i+1}} + \frac{1}{a_i^r a_{i+1}^r a_{i+2}} + \cdots . $$

We have $z_{i+1} = (z_i a_i - 1) a_i^{r-1}$, and $0 < z_i < 1$.
Suppose that $z_1$ is a rational number with denominator $q$. Then all $z_i$ can be written with that same denominator. But there are finitely many such fractions between 0 and 1, so we must have

$$ z_j = z_{j+k} = \frac{p}{q} ~~\text{ for some } j, k > 0. $$

Then

$$\begin{align} 0 &< z_{j+1} = \left(\frac{p a_j}{q} - 1\right) a_j^{r - 1} < 1, ~~\text{ and } \\ 0 &< z_{j+k+1} = \left(\frac{p a_{j+k}}{q} - 1\right) a_{j+k}^{r - 1} < 1. \end{align}$$

Observe that if

$$\begin{align} 0 < \left( \frac{pa}{q} - 1 \right) a^{r-1}, \\ \left( \frac{p(a+1)}{q} - 1 \right) (a+1)^{r-1} < 1. \end{align}$$

Consequently, we have $q < pa$, and

$$\begin{align} &\quad \frac{p}{q} (a^{r-1} + 1) + \left(\frac{pa}{q} - 1\right) \\ &< \frac{p}{q} (a^{r-1} + 1) + \left(\frac{pa}{q} - 1\right) + \left( \frac{pa}{q} - 1 \right) a^{r-1} \\ &= \left( \frac{p(a+1)}{q} - 1 \right)(a^{r-1}+1) \\ &< \left( \frac{p(a+1)}{q} - 1 \right)(a+1)^{r-1} < 1, \end{align}$$

i.e.,

$$\begin{align} p(a^{r-1} + 1) + (pa - q) < q < pa. \end{align}$$

Hence,

$$\begin{align} p(a^{r-1} + 1) < q < pa, \end{align}$$

which is a contradiction. Therefore, $a_{j+k} = a_j$.

At last, a simple induction will imply that $a_{i+k} = a_i$ for all $i \ge j$.