Given that $\left(\frac{a^2}{2}a_n-ab_n\right)_n$ converges for all $a\in(0,1]$, show that $(a_n)_n,(b_n)_n$ are convergent

Question

We have 2 sequences $(a_n)_n,(b_n)_n$. It is known that the sequence $\left(\frac{1}{2}a_n-b_n\right)_n$ is convergent and $\left(\frac{a^2}{2}a_n-ab_n\right)_n$ converges for all $a\in (0,1]$. Can somebody help me how to prove that $(a_n)_n,(b_n)_n$ are convergent? Please

Answer 1

Since $\frac{a^2}{2}a_n-ab_n$ converges for $a \in (0,1]$, also $\frac{1}{a}\cdot (\frac{a^2}{2}a_n-ab_n) = \frac{a}{2}a_n-b_n$ converges. Hence $(\frac{a}{2}a_n-b_n) - (\frac{1}{2}a_n-b_n) = \frac{a-1}{2}a_n$ converges. We conclude that $\frac{2}{a-1} \cdot \frac{a-1}{2}a_n = a_n$ converges (note $a-1 \ne 0$ for $a < 1$). Hence also $\frac{1}{2}\cdot a_n - (\frac{1}{2}a_n-b_n) = b_n$ converges.

Answer 2

Let $f_n(a):=\tfrac{a^2}{2}a_n-ab_n$. Given that $(f_n(a))_n$ converges for all $a\in(0,1]$, so do $$(4f_n(1)-8f_n(\tfrac12))_n=(a_n)_n \qquad\text{ and }\qquad (f_n(1)-4f_n(\tfrac12))_n=(b_n)_n.$$