Winner: Zachary Li

Runner-Up: Eric Peng

Correct solutions were also submitted by Ria Garg and Victor Lin

Winner: Victor Lin

Runner-Up: Tyson Ramirez

A correct solution was also submitted by Ria Garg

The ${\it Bell \ Numbers \ }$, $b_n$, are the number of ways of partitioning a set of $n$ objects into subsets.

We define $b_0=1$. It can be shown that $b_1=1, b_2=2, b_3=5, b_4 = 15, $ and more generally:

$$ b_{n+1} = \sum\limits_{k=0}^{n} {n \choose k} b_k. $$

Prove that:

$$ e^{\left({ e^x-1}\right)} = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \ \ \textrm{ and } \ \ b_n = \frac{1}{e} \sum_{k=1}^{\infty} \frac{k^n}{k!}. $$

Winner: Eric Peng

Runner-Up: Victor Lin

Winner: Victor Lin

Runner-Up: Tyson Ramirez

Correct solutions were also submitted by: Merdangeli Bayramov, Ria Garg, and Eric Peng.

Winner: Saisneha Ghatti

Runner-Up: Zachary Li

Correct solutions were also submitted by Ria Garg, Helen Li, and Victor Lin.

Winner: Victor Lin

Runner-Up: Zachary Li

A correct solution was also submitted by Saisneha Ghatti and Eric Peng.

Winner: Eric Peng

Runner-Up: Victor Lin

Correct Solutions were also submitted by Ria Garg, Saisneha Ghatti, Matthew Li

Winner: Victor Lin

Runner-Up: Saisneha Ghatti

A correct solutions was also submitted by Eric Peng

Partial solutions were submitted by Atharv Chagi, Ria Garg, Zachary Li

Winner: Eric Peng

Runner-Up: Ritwik Chaudhuri

A correct solution was also submitted by Ria Garg

Determine:

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

Winner: Eric Peng

Runner-Up : Zachary Li

A correct solution was also submitted by Atharv Chagi

HINT: It is known that if $A$ is the area and $P$ the perimeter of the triangle then $A=\frac{1}{2} rP$.

Determine:

$$ \sum_{n=1}^{\infty} \left( \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} \right) \frac{1}{n}. $$

(You may assume this series converges).

Hint: First show:

$$ \frac{1}{\sqrt{1-4x}} = 1 + \sum_{n=1}^{\infty} { 2n \choose n} x^n \textrm{ for } |x|< \frac{1}{4} $$

and then:

$$ 2 \ln\left( \frac{1 - \sqrt{1-4x}}{2x} \right) = \sum_{n=1}^{\infty} \frac{1}{n} { 2n \choose n} x^n \textrm{ for } |x|< \frac{1}{4}. $$

Winner: Eric Peng

Runner-Up: Nicolas Heredia

HINT: It is known that if $A$ is the area of the triangle and $P$ the perimeter then $A= \frac{1}{2} rP.$

Winner: Eric Peng

Runner-Up: James Heath

Determine:

\begin{equation} \prod_{n=2}^{\infty} \frac{n^3-1}{n^3 +1}. \end{equation}

Note: $\prod\limits_{n=2}^{\infty} a_n= \lim\limits_{n \to \infty} (a_2 a_3 \cdots a_n).$

Winner: Michalis Paizanis

Runner-Up: Eric Peng

Let $0<b_1<a_1$ and let:

$$ a_{n+1} = \frac{a_{n} + b_{n}}{2} \textrm{ and } b_{n+1} = \frac{2a_n b_n}{a_n + b_n}. $$

Show that: $$ 0 < b_n < b_{n+1} < a_{n+1} < a_n $$ and determine:

$$ \lim_{n \to \infty} a_n \textrm{ and } \lim_{n \to \infty} b_n. $$

Winner: Eric Peng

Runner-Up: James Heath

A correct solution was also submitted by Michalis Paizanis

Let $a>0$ and $d>0.$
Let: $$A_{n} = \frac{1}{n}\sum_{k=0}^{n-1} (a + kd)$$
and: $$ G_{n} = \left( a(a+d)(a+2d)\cdots(a+(n-1)d) \right)^{1/n}. $$

Determine:
$$ \lim_{n \to \infty} \frac{G_n}{A_{n}}.$$

Winner: Eric Peng

Runner-Up: Yugendra Uppalapati

A correct solution was also submitted by Michalis Paizanis

Sept. 2021

Determine: $$ \lim_{n \to \infty} n \sin(2\pi e n!). $$

Hint: $ e =1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots. $

Winner: Michael Holland,

Runner-Up: Divya Darji

Find a simple formula for:

$$ 1{n \choose 1} + 2 { n \choose 2} + 3 {n \choose 3} + \cdots + n{ n \choose n}. $$

Show that there is a continuous function $f: {\mathbb R} \to {\mathbb R}$ that takes on every value exactly 3 times. (That is, for any given $y_0 \in {\mathbb R}$ there are exactly three solutions $x_1 < x_2 < x_3$ of $f(x)=y_0$).

Note: It is known that there is no continuous function $f: {\mathbb R} \to {\mathbb R}$ that takes on every value exactly 2 times.

Feb. 2021

Show that in any Pythagorean triangle one of the sides must be divisible by 3, one side must be divisible by 4, and one side must be divisible by 5.
(A ''Pythagorean triangle'' is a right triangle with sides $a,b$ and hypotenuse $c$ such that $a,b,c$ are positive integers and $c^2=a^2+b^2$).

Note: A Pythagorean triple $(a,b,c) $ is called ''primitive'' if $(a,b,c)$ have no common factors. It is known that every primitive Pythagorean triple can be written in the form $a=m^2-n^2, b= 2mn, c = m^2 + n^2$ where $m$, $n$ are positive integers with $m>n.$

Hint: Try to prove the result for primitive Pythagorean triples first.

Winner: Michalis Paizanis

Runner-Up: Angela Yuan

a correct solution was also submitted by Neel Shanmugam

Winner: Michalis Paizanis

Runner-Up: Subiksha Sankar

Winner: Michalis Paizanis

Runner-Up: Angela Yuan

other correct solutions by: Alejandro Castellanos, Subisksha Sankar

Winner: Subiksha Sankar

Runner-Up: Michalis Paizanis

Winner: David Duhon

Runner Up: Peyton Thibodeaux

other correct solutions by: Michalis Paizanis, Subiksha Sankar

Show that: $$(\sin^{-1}(x))^2 = \sum_{n=0}^{\infty} \frac{2^{2n} (n!)^2 \, x^{2n+2}}{(2n+1)! (n+1)} $$ on $[0,1].$

Use this to determine: $$ \sum_{n=1}^{\infty} \frac{1}{ n^2 {2n \choose n} } . $$

Hint: Let: $$y = \sum_{n=0}^{\infty} \frac{2^{2n} (n!)^2 \, x^{2n+2}}{(2n+1)! (n+1)} $$ and show that $y$ satisfies a second order linear differential equation. Then solve the differential equation.

Winner: Jingyi Dai

Runner-Up: Hubert Yang

Other correct solutions by: Anush Beeram, Elson Darby, David Duchon, Rhythm Garg Harrison Rodi

Winner: David Duhon

Runner-Up: Rhythm Garg

Find all functions $f: {\mathbb R} \to {\mathbb R}$ that satisfy: $f(x+y) = f(x) + f(y)$ and $f(xy)=f(x)f(y).$ (You may NOT assume that $f$ is continuous!) Hint: Try to show $f$ is nondecreasing.

Winner: Rhythm Garg

Prove the following:

$$ \frac{1}{3} = \frac{1+3}{5+7} = \frac{1+3+5}{7+9+11} = \cdots. $$

Winner: Chris Howard

Runner-Up: Andrew Sansom

Winner: Arnav Iyer

Runner-Up: Aryan Agarwal

Let $0<x<1$. Simplify as much as possible:

$$ \ln(x) \ln(1-x) + \sum_{n=1}^{\infty} \frac{x^{n}}{n^2} +\sum_{n=1}^{\infty} \frac{(1-x)^{n}}{n^2}. $$

Winner: Weilun Sun

Runner-Up: Hubert Yang

Sept. 2019

Let $a$ and $b$ be real numbers with $b\neq 0$. Determine:

$$ \int_{0}^{2\pi} \frac{d\theta}{\cos^{2}(\theta) + 2a\cos(\theta)\sin(\theta) + (a^{2}+b^{2})\sin^{2}(\theta)}. $$

Apr. 2019

Determine all positive integer solutions to the system of equations:

$$ xy=z+w$$

$$ zw=x+y.$$

Winner: Rhythm Garg

Runner-Up: Jingyi Dai

Winner: Edoardo Luna

Runner-Up: Rhythm Garg

Take a square piece of paper with vertices labeled $A, B, C, D$ when proceeding in the counterclockwise direction with vertex $A$ in the upperleft corner. Now take the side $AB$ and fold it in such a way that $A$ gets folded to a point $A'$ and $B$ to $B'$ where $B'$ is on $CD$ and $A'$ is outside the square $ABCD$. Then $A'B'$ intersects $AD$ at $E$. Next draw a circle that is tangent to $CD$, tangent to $AD$ and tangent to $A'B'$. Denote the radius of this circle by $r$. Prove that $r= A'E.$

Hint: A well-known fact from geometry states that $r= \frac{2 \textrm{Area}(T)}{\textrm{Perimeter}(T)} $ where $T$ is the triangle $EB'D$.

Winner: Rhythm Garg

Runner-Up: Fernando Moreno

Winner: Rhythm Garg

Runner-Up: Trenton Hicks

Let $0<a<b$. Determine:

$$ \int_{a}^{b} \cos^{-1}\left( \frac{x}{\sqrt{(a+b)x - ab}} \right)\, dx. $$

Winner: Tien Le

Runner-Up: Rhythm Garg

Investigate the convergence of:

$$ \sum_{n=3}^{\infty} \frac{1}{\ln(\ln(n))^{\ln(n)}} $$

and:

$$ \sum_{n=3}^{\infty} \frac{1}{\ln(\ln(n))^{\ln(\ln(n))}} $$

Let $a>0$ and $b>0$. Determine:

$$ \int_{0}^{2\pi} \frac{ab}{a^2 \cos^{2}(t) + b^2\sin^{2}(t)} \, dt. $$

Determine:

$$ \sin\left(\frac{\pi}{n}\right) \sin\left(\frac{2\pi}{n}\right) \sin\left(\frac{3\pi}{n}\right) \cdots \sin\left(\frac{(n-1)\pi}{n}\right). $$

Hint: Note that: $$ z^{n}-1 = (z-1)(z - e^{\frac{2\pi i}{n}}) (z - e^{\frac{4\pi i}{n}}) (z - e^{\frac{6\pi i}{n}}) \cdots (z - e^{\frac{2(n-1)\pi i}{n}}). $$

Let $x>1$ and define $f(e)=e.$ For $x>1$ and $x\neq e$ let $f(x)>0$ be the unique number such that $f(x)\neq x$ and $x^{f(x)} = f(x)^{x}.$Sketch the graph of $f(x)$ and calculate $f'(x)$ (you may leave this answer in terms of $x$ and $f(x)$). Finally calculate $f'(e)$.

Jan. 2018

Let $a, b, c$ be real numbers such that $ax^{2} + 2bxy +cy^2>0$ for all $(x,y)\neq (0,0)$. Determine: $$ \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \left( e^{-(ax^2 + 2bxy + cy^2)} \right) \, dx \, dy.$$ You may use the well-known fact that for $H>0$: $$ \int_{-\infty}^{\infty} e^{-Ht^2} \, dt = \sqrt\frac{\pi}{H}. $$

Determine: $$ \sum_{n=2}^{\infty} (-1)^{n}\frac{\ln(n)}{n}. $$ Express your answer in terms of Euler's constant, $\gamma$.Note that: $\gamma = \lim\limits_{n \to \infty} \left( 1 + \frac{1}{2} + \cdots \frac{1}{n} - \ln(n) \right). $Hint: You may assume there exists a constant $A$ such that: $$ \lim\limits_{n \to \infty} (\frac{\ln(2)}{2} + \frac{\ln(3)}{3} + \cdots + \frac{\ln(n)}{n} - \frac{1}{2}\ln^{2}(n) ) = A. $$

Winner: Linda Yu

Runner-Up: Austin McGregor

Winner: Brandon Ohl

Runner-Up: Ethan Seal

Consider the sum:
$$ \begin{array}{1 2 3 4 5 6 7 }
\ & \ & \textrm{M} & \textrm{A} & \textrm{J} & \textrm{O} & \textrm{R} \\
\ & + & \textrm{M} & \textrm{I} & \textrm{N} & \textrm{O} & \textrm{R} \\
= & \textrm{R} & \textrm{E} & \textrm{S} & \textrm{U} & \textrm{M} & \textrm{E} \end{array}
$$

where each letter represents a different nonnegative integer 0, 1, .... 9. What is the largest number that RESUME can represent and still have the equation be true?

Note: There are 10 different letters in this sum so all of the nonnegative integers from 0 to 9 will be used exactly once.

Winner: Ethan Seal

Runner-Up: Brandon Ohl

Let $A=(0,2)$, $B = (3,0)$. Find a point $C$ on the unit circle centered at the origin so that triangle $ABC$ is of largest area.

Winner: William Liu

Runner-Up: Xiangyu Kong

November 2016

Winner: William Liu

Runner-Up: Aoxue Chen

October 2016

a. Let $A$ and $B$ be two linear transformations from ${\mathbb R}^N \to {\mathbb R}^N.$ Show that it is impossible for $AB-BA=I$ where $I$ is the identity map.

b. On the other hand show that it is possible to find two linear transformations (defined on infinite dimensional spaces) with $AB-BA=I$.

Winner: William Liu

Determine $$ \lim_{n \to \infty} \frac{1}{n^4} \prod_{i=1}^{2n} (n^2 + i^2)^{\frac{1}{n}}. $$

Note: The $\prod$ sign is a short-hand for product. That is: $\prod\limits_{i=1}^{2n} b_i = b_{1}b_{2} \cdots b_{2n-1} b_{2n}.$

Winner: William Liu

Runner-Up: Christopher Lee

Winner: William Liu

Runner-Up: Jagath Vytheeswaran

$$ \min_{p} \max_{x \in [a,b]} \frac{|x-p|}{x}. $$

Winner: William Liu

Runner-Up Brandon Ohl

Let $f_{n}$ be the Fibonacci sequence. Determine $$\sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{f_{2n+1}} \right). $$

Hint: Use a trig identity and the following identity which holds for the Fibonacci sequence: $f_{n+1}f_{n+2} -f_{n}f_{n+3} = (-1)^n. $

Determine all nonnegative continuous functions which satisfy:

$$ f(x+t) = f(x) + f(t) + 2 \sqrt{f(x)}\sqrt{f(t)} \textrm{ for } x \geq 0, t \geq 0. $$

December 2014

Winner: Murray Lee

Runner Up: Steven Grigsby

Winner: Steven Grigsby

Runner Up: Mark Fincher

September 2014

Simplify: $$ \frac{1}{\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{4}} + \frac{1}{\sqrt[3]{4} + \sqrt[3]{6}+\sqrt[3]{9}} + \frac{1}{\sqrt[3]{9} + \sqrt[3]{12} + \sqrt[3]{16}} $$

Winner: Tamaki Ueno

Runner Up: Murray Lee

April 2014

Squared Matrices and Characteristic Polynomials. Let $A$ be a real $4\times4$-matrix with characteristic polynomial $p(\lambda)=\det(\lambda{I}-A)=\lambda^4-s_1\lambda^3+s_2\lambda^2-s_3\lambda+s_4.$ Show that if $A$ admits real square roots, in the sense that there is a real $4\times4$-matrix $S$ such that $S^2=A,$ then $s_1+s_2+s_3+s_4\ge-1.$

Winner: Mark Fincher

Runner-Up: Chia-Ting Han

Winner: Kevin Lin

Runner-Up: Chia-Ting Han

Winner: Chia-Ting Han

Runner-Up: Kevin Lin

Winner: Chia-Ting Han

Runner-Up: Kevin Lin

Winner: Kevin Lin

Runner-Up: Chris James

Integration Bee Warm-Up. Find an anti-derivative of the function $$f(x)=\sqrt{\sqrt{\sqrt{x}+1}-\sqrt{\sqrt{x}-1}},\;x>1.$$

Winner: Heather Olney

Runner-Up: Alyssa Sylvester