Determine the integer part

Question

Let $(a_n)$ be a sequence given by $a_1=\frac{1}{2}$ and $a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}$ , $n=1,2,...$ Let $b_n=a_1+a_2+...+a_n$ for each positive integer $n$

Determine the integer part $[b_n]$

Answer 1

Let's start by showing that $b_n = 2-\frac{1}{1-a_{n+1}}$. Proof by induction:

• $n=0$: $0 = 2-\frac{1}{1-\frac{1}{2}}=0$. Check.
• Going from $(n-1)$ to $n$: $$\begin{eqnarray}b_n = b_{n-1}+a_n & = & 2-\frac{1}{1-a_n} + a_n\\ & = & 2 - \frac{1-a_n+a_n^2}{1-a_n}\\ & = & 2 - \frac{1}{1-\frac{a_n^2}{a_n^2-a_n+1}}\\ & = & 2 - \frac{1}{1-a_{n+1}}\end{eqnarray}$$

The other ingredient is the inequality $0<a_n\leq \frac{1}{2}$, provable by induction as well:

• $n=1$: $0<\frac{1}{2}\leq \frac{1}{2}$. Check.
• Going from $n$ to $(n+1)$: $$\begin{eqnarray} 0 < a_n \Rightarrow 0 < a_n^2-a_n+1 \Rightarrow 0 < \frac{a_n^2}{a_n^2-a_n+1} = a_{n+1} \\ a_n \leq \frac{1}{2} \Rightarrow a_n^2 \leq \frac{1}{4} \wedge \frac{1}{2} \leq a_n^2 - a_n + 1 \Rightarrow a_{n+1} = \frac{a_n^2}{a_n^2-a_n+1} \leq \frac{1}{2}\end{eqnarray}$$

Combining these two observations tells us that $0\leq b_n<1$, which implies $\lfloor b_n\rfloor = 0$.