Unlike many of the unsuccessful internet searches, two months ago I came across the professional website of Professor Roberto Tauraso from the Mathematics Department of the Università di Roma. He dedicated one section of his website to solve and share the American Monthly Mathematical Problems, which are part of The American Mathematical Monthly. To be honest, I was not familiar with this journal, however I decided to give it a try and attempt a few problems.

I have always been a fan of such math problems that push you to find interesting approaches and learn new obscure theorems in order to derive a solution. The problems I found were precisely like that. I must admit some of them took me several days and a few cups of coffee, yet I take pride to say I was not totally lost, after all. These are the problems I have solved so far (will add more as time goes on). I would invite the readers to take them as sporadic challenges.

Problem 12260

Prove

\[\int_{0}^\infty \frac{\sin^2(x) - x \sin(x)}{x^3} dx = \frac12 - \log2\]

Proposed by S. M. Stewart (Australia), American Mathematical Monthly, Vol.128, 2021

Solution

Problem 12265

For a fixed positive integer \(k\), let \(a_0 = a_1 = 1\) and \(a_n = a_{n-1} + (k-n)^2 a_{n-2}\) for \(n \geq 2\). Show that \(a_k = (k-1)!\). Proposed by R. Dempsey (USA), American Mathematical Monthly, Vol.128, 2021

Solution

Problem 12276

Prove

\[I = \sum_{n=2}^\infty \frac{1}{n+1} \sum_{i=1}^{\lfloor \frac{n}{2} \rfloor} \frac{1}{2^{i-1}(i-1)!(n-2i)!} = 1\]

Proposed by Joe Santmyer, Las Cruces, NM., Vol 128, Issue 8, American Mathematical Monthly, 2021.

Solution

Problem 12279

Let \(j\) and \(n\) be positive integers of opposite parity with \(j<n\). Prove

\[\sum_{k=j}^n \frac{(-1)^k (k-1)!}{2^k} \binom{k}{j} {n \brace k} = 0\]

where \({n \brace k}\) are the Stirling numbers of second kind. Proposed by B. Isaacson (USA), American Mathematical Monthly, 2021.

Solution

Problem 12287

Prove

\[\sum_{n=1}^\infty \left( n \left( \sum_{k=n}^\infty \frac{1}{k^2} \right)^2 - \frac{1}{n} \right) = \frac12\left( 3 - \zeta(2) + 3\zeta(3) \right)\]

Proposed by - O. Furdui and A. Sintamarian (Romania), Vol 128, American Mathematical Monthly 2021.

Solution

Problem 12300

Let \(n\) be an integer such that \(n \geq 3\). Prove that there is no permutation \(\pi\) of \(\{ 1, 2, \cdots, n \}\) such that \(\pi(1), 2\pi(2), \cdots, n\pi(n)\) are distinct modulo \(n\). Proposed by H. A. ShahAli, Tehran, Iran. American Mathematical Monthly, 2022.

Solution

Problem 12316

For each \(i\) in \(\{1, 2, \cdots C\}\) , we have \(2i\) coins with color $i$. Place these \(C(C+1)\) coins in a line. A move consists of the transposition of two adjacent coins. Let \(m\) be the minimum number of moves required to reach a configuration where all coins of the same color are together in a run of consecutive coins. Show that the maximum value of $m$ over all initial configurations is \((C-1)C(C+1)(3C+2)/12\). Proposed by H. A. ShahAli, Tehran, Iran, and Manija Shahali, Bakersfield, CA. American Mathematical Monthly, Vol 129 No 4, 2022.

Problem 12349

Let \(A_n\) be the set of permutations of \(\{ 1, \cdots, n \}\) that have at least one fixed point. For \(\pi \in A_n\), we write \(\text{Fix}(\pi)\) for \(\{ j : \pi(j) = j\}\). Evaluate \(S = \sum_{\pi \in A_n} \left( \frac{\text{sign}(\pi)}{| \text{Fix}(\pi) |} \sum_{j \in \text{Fix}(\pi) } j \right)\). Proposed by Roberto Tauraso, Tor Vergata University of Rome, Italy. American Mathematical Monthly, Vol 129 No 8, 2022.

Problem 12377

An integer is a one-drop number if its decimal digits \(d_1, \cdots, d_n\) satisfy \(1 \leq d_1 \leq \cdots \leq d_i > d_{i+1} \leq \cdots \leq d_n\) for some \(i\). For \(n\geq 2\), how many \(n\)-digit one drop numbers are there? Proposed by Li Zhou, USA. American Mathematical Monthly, 2023.

Problem 12385

Let $n$ be a positive integer. Prove

\[\sum_{1\leq i \leq k \leq n } \frac{(-2)^k}{k+1} \binom{n}{k} \binom{k}{i}^{-1} = \frac{(-1)^n - 1}{2n}\]

Proposed by H. Ohtsuka, Japan. American Mathematical Monthly, 2023.

More solutions will be available after the deadline.

One of the main intentions of this post is to thank Professor Tauraso for sharing knowledge, motivating with his work, and freely exchanging ideas with students across the internet. If someday I visit Rome again, I will definitely pass by your office to meet you. Until then, Grazie mille! :)