Mathematical Reflections 2015, Issue 1 - Problem U327

Problem:
Let $(a_n)_{n \geq 0}$ be a sequence of real numbers with $a_0=1$ and $$a_{n+1}=\dfrac{a_n}{n^2a_n+a^2_n+1}.$$
Find the limit $\displaystyle \lim_{n \to \infty} n^3 a_n$.

Proposed by Khakimboy Egamberganov.

Solution:
Observe that $a_{n+1}<\dfrac{a_n}{n^2a_n}=\dfrac{1}{n^2}$, so $\displaystyle \sum_{n=0}^{\infty} a_n$ converges by the Comparison Test. Let $$\sum_{n=0}^{\infty} a_n=c \in \mathbb{R}.$$
We have $$\dfrac{1}{a_{n+1}}=n^2+a_n+\dfrac{1}{a_n}.$$ So, $$\begin{array}{rcl} \dfrac{1}{a_1}&=&0^2+a_0+\dfrac{1}{a_0} \\ \dfrac{1}{a_2}&=&1^2+a_1+\dfrac{1}{a_1} \\ \vdots & \vdots & \vdots \\ \dfrac{1}{a_n}&=&(n-1)^2+a_{n-1}+\dfrac{1}{a_{n-1}},\end{array}$$ and summing the two columns, we get $$\dfrac{1}{a_n}=\dfrac{(n-1)n(2n-1)}{6}+\sum_{k=0}^{n-1} a_k+1.$$
Hence, $$\dfrac{1}{n^3a_n}=\dfrac{(n-1)n(2n-1)}{6n^3}+\dfrac{\sum_{k=0}^{n-1} a_k+1}{n^3}.$$ Since $$\displaystyle \lim_{n \to \infty} \dfrac{\sum_{k=0}^{n-1} a_k+1}{n^3}=\lim_{n \to \infty} \dfrac{c+1}{n^3}=0,$$ we get $$\lim_{n \to \infty} \dfrac{1}{n^3 a_n}=\lim_{n \to \infty} \dfrac{(n-1)n(2n-1)}{6n^3}=\dfrac{1}{3},$$ so $$\lim_{n \to \infty} n^3 a_n=3.$$

Mathematical Reflections 2015, Issue 1 - Problem U326

Problem:
Find $$\sum_{n=0}^\infty \dfrac{a_n+2}{a^2_n+a_n+1},$$ where $a_0>1$ and $a_{n+1}=\dfrac{1}{3}(a^3_n+2)$ for all integers $n \geq 0$.

Proposed by Arkady Alt.

Solution:

Since $a_0>1$, then it can be proved by induction that $a_n>1$ for all $n \in \mathbb{N}$.
For all $n \geq 0$ we have $$\begin{array}{lll} \dfrac{a_n+2}{a^2_n+a_n+1}&=&\dfrac{(a_n+2)(a_n-1)}{a^3_n-1}\\ &=& \dfrac{(a_n+2)(a_n-1)}{3(a_{n+1}-1)}\\ &=&\dfrac{a^2_n+a_n+1-3}{3(a_{n+1}-1)}\\ &=&\dfrac{\frac{a^3_n-1}{a_n-1}-3}{3(a_{n+1}-1)} \\  &=& \dfrac{\frac{3(a_{n+1}-1)}{a_n-1}-3}{3(a_{n+1}-1)} \\ &=&\dfrac{a_{n+1}-a_n}{(a_n-1)(a_{n+1}-1)}\\&=&\dfrac{1}{a_n-1}-\dfrac{1}{a_{n+1}-1}  \end{array}$$
Therefore, $$\sum_{n=0}^\infty \dfrac{a_n+2}{a^2_n+a_n+1}=\dfrac{1}{a_0-1}.$$

Mathematical Reflections 2015, Issue 1 - Problem J325

Problem:
For positive real numbers $a$ and $b$, define their \emph{perfect mean} to be half of the sum of their
arithmetic and geometric means. Find how many unordered pairs of integers $(a,b)$ from the set $\{1,2,\ldots,2015\}$ have their perfect mean a perfect square.

Proposed by Ivan Borsenco.

Solution:

We have $$\dfrac{\frac{a+b}{2}+\sqrt{ab}}{2}=n^2 \qquad n \in \mathbb{N},$$ i.e. $$\sqrt{a}+\sqrt{b}=2n, \qquad n \in \mathbb{N}.$$ As $a,b \in \{1,2,\ldots,2015\}$, then $a$ and $b$ must be perfect squares and $n \leq 44$. Assume that $a \leq b$. If $a$ is an odd perfect square, then $b$ is an odd perfect square, and making a case by case analysis for $a \in \{1,3^2,\ldots,43^2\}$ we get $22+21+\ldots+2+1=253$ unordered pairs. Similarly, if $a$ is an even perfect square, then $b$ is an even perfect square and we get $22+21+\ldots+2+1=253$ unordered pairs. So, there are $253+253=506$ unordered pairs which satisfy the given conditions.