Convergence of Sequences Given a Quadratic Sum Condition Let ( u n ) (u_n) ( u n ) ( v n ) (v_n) ( v n )
u n 2 + u n v n + v n 2 → n → + ∞ 0. u_n^2+ u_n v_n + v_n^2 \xrightarrow[n\to+\infty] {}0. u n 2 + u n v n + v n 2 n → + ∞ 0.
Prove that the sequences ( u n ) (u_n) ( u n ) ( v n ) (v_n) ( v n )
Show Answer:
We have
0 ≤ ( u n + v n ) 2 = u n 2 + 2 u n v n + v n 2 ≤ 2 ( u n 2 + u n v n + v n 2 ) → n → + ∞ 0. 0\le (u_n+v_n)^2= u_n^2+2u_n v_n+ v_n^2\le 2(u_n^2+u_n v_n+ v_n^2)\xrightarrow[n\to+\infty]{}0. 0 ≤ ( u n + v n ) 2 = u n 2 + 2 u n v n + v n 2 ≤ 2 ( u n 2 + u n v n + v n 2 ) n → + ∞ 0.
Hence
u n + v n → n → + ∞ 0 u_n+v_n \xrightarrow[n\to+\infty]{}0 u n + v n n → + ∞ 0 . Also
u n v n = ( u n + v n ) 2 − ( u n 2 + u n v n + v n 2 ) → n → + ∞ 0 u_n v_n= (u_n+v_n)^2 -(u_n^2+u_n v_n+ v_n^2)\xrightarrow[n\to+\infty]{}0 u n v n = ( u n + v n ) 2 − ( u n 2 + u n v n + v n 2 ) n → + ∞ 0
so
u n 2 + v n 2 = 2 ( u n 2 + u n v n + v n 2 ) − ( u n + v n ) 2 → n → + ∞ 0 u_n^2+v_n^2= 2(u_n^2+u_n v_n+ v_n^2)-(u_n+v_n)^2\xrightarrow[n\to+\infty]{}0 u n 2 + v n 2 = 2 ( u n 2 + u n v n + v n 2 ) − ( u n + v n ) 2 n → + ∞ 0
which allows to conclude that
u n → n → + ∞ 0 u_n\xrightarrow[n\to+\infty]{}0 u n n → + ∞ 0 and
v n → n → + ∞ 0 v_n\xrightarrow[n\to+\infty]{}0 v n n → + ∞ 0 .