Question 3.T.4: Let (xn) and (yn) be two convergent sequences with limits x ...
Let (x_{n}) and (y_{n}) be two convergent sequences with limits x and y, respectively. Then
(i) (x_{n} + y_{n}) converges to x + y,
(ii) (x_{n}y_{n}) converges to to xy,
(iii) (x_{n}/y_{n}) converges to x / y, provided y ≠ 0.
(i) Since x_{n} and y_{n} are convergent, given ε > 0, there are integers N_{1} and N_{2} such that
n ≥ N_{1} ⇒ |x_{n} − x| < ε
n ≥ N_{2} ⇒ |y_{n} − y| < ε.
Let N = max\left\{N_{1}, N_{2}\right\}. If we take n ≥ N then both inequalities n ≥ N_{1} and n ≥N_{2} are satisfied, and it follows, using the triangle inequality, that
|(x_{n} + y_{n}) − (x + y)| = |(x_{n} − x) + (y_{n} − y)|
≤ |x_{n} − x| + |y_{n} − y|
≤ 2ε.
Hence (see Remark 3.1.4) lim(x_{n} + y_{n}) = x + y.
(ii) Let ε > 0. For any n ∈\mathbb{N}, we have
|x_{n}y_{n} − xy| = |x_{n}y_{n} − x_{n}y + x_{n}y − xy|
≤ |x_{n}y_{n} −x_{n}y| + |x_{n}y − xy|
= |x_{n}| |y_{n} − y| + |y| |x_{n} − x| . (3.4)
Since the sequence (x_{n}) is convergent, it is bounded (by Theorem 3.2), hence there is a positive constant K such that
|x_{n}| ≤ K for all n ∈ \mathbb{N}.
Taking the integers N_{1}, N_{2}, N as in part (i), we have, in view of (3.4),
n ≥ N ⇒ |x_{n}y_{n} − xy| ≤ K |y_{n} − y| + |y| |x_{n} − x|
< Kε + |y| ε = cε,
where c = K + |y| . This proves that x_{n}y_{n} → xy.
(iii) Since y_{n} → y ≠0, we know from Theorem 3.3 that there is a positive constant M and an integer N_{0} such that
|y_{n}| > M for all n ≥ N_{0}.
Consequently x_{n}/y_{n} is defined for all n ≥ N_{0}, and the limit of (x_{n}/y_{n}) is taken to mean the limit of the tail (x_{n}/y_{n} : n ≥ N_{0}). We have
\left|\frac{x_{n}}{y_{n}} − \frac{x}{y} \right| = \frac{\left|x_{n}y − xy_{n}\right| }{\left|y_{n}\right|\left|y\right| }
≤\frac{\left|x_{n}y − xy\right| + \left|xy − xy_{n}\right|}{\left|y_{n}\right|\left|y\right|}
=\frac{\left|x_{n} − x\right| }{\left|y_{n}\right|} + \frac{\left|x\right| \left|y − y_{n}\right| }{\left|y_{n}\right|\left|y\right|}. (3.5)
Suppose we are given a positive ε. As in (i), there are integers N_{1} and N_{2} such that |x_{n} − x| < ε for all n ≥ N_{1} and |y_{n} − y| < ε for all n ≥ N_{2}. Taking N = max\left\{N_{0}, N_{1}, N_{2}\right\} , and recalling (3.5), we conclude that
n ≥ N ⇒ n ≥ N_{0}, n ≥ N_{1}, n ≥ N_{2}
⇒ \left|\frac{x_{n}}{y_{n}} − \frac{x}{y} \right| < \frac{ε}{M} + \frac{\left|x\right| }{\left|y\right| M}ε = cε,
where c = \frac{1}{M} + \frac{\left|x\right| }{\left|y\right| M}.