# Thread: two sequences converging to 0

1. ## two sequences converging to 0

Let (an) and (bn) be two sequences converging to 0. Prove that the sequence (anbn) converges to 0. (Must prove the statement directly from the definition of convergence.)

I don't know how to prove by definition of convergence.I'm a bit confused with "definition of limit" and "definition of convergence". Are they same which is the "epsilon-delta definition".

2. I believe that it is the same as the epsilon delta proof. Basically, given epsilon > 0, you need to find an N such that n> N implies the magnitude of an times bn is < epsilon. To do this you need to use the epsilon-delta definitions of an and bn's convergences.

3. $\displaystyle \lim_{n\to\infty} a_n = 0$
$\displaystyle \lim_{n\to\infty} b_n = 0$

$\displaystyle \forall \epsilon > 0 \exists n_\epsilon \in N: \forall n'\ge n_\epsilon | a_n' - 0 | < \epsilon$ same holds for $\displaystyle \b_n$

$\displaystyle \ |a_n\cdot b_n - 0 | = |a_n||b_n|$
since $\displaystyle \forall \epsilon > 0 \exists n_\epsilon \in N: \forall n'\ge n_\epsilon | a_n' - 0 | < \epsilon$ set $\displaystyle \epsilon = \sqrt\epsilon '$
so $\displaystyle \exist n_\epsilon : \forall n' \ge \n_\epsilon |a_n| < \epsilon = \sqrt\epsilon '$ same for b_n, lets call the resulting $\displaystyle \ n_\epsilon$ for $\displaystyle \ b_n m_\epsilon$

Let $\displaystyle \ k = \max ( n_\epsilon , m_\epsilon )$
so $\displaystyle \forall n'\ge k |a_n'\cdot b_n | = |a_n|\cdot |b_n| < \ (\sqrt\epsilon ' )^2 = \epsilon'$

or you can use the fact that all 1) sequences converge if and only if they are cauchy in R (a complete space) and 2) all cauchy sequences are bounded to prove that. use the identity
$\displaystyle \ |a_n\cdot b_n - a_m\cdot\b_m| = |a_n\cdot b_n - a_m\cdot b_n + b_n\cdot a_m - a_m\cdot b_m| = |(a_n\cdot b_n - a_m\cdot b_n) + (b_n\cdot a_m + b_m\cdot a_m) |$
by triangle inequality this simplifies too $\displaystyle \ | a_n\cdot b_n - a_m\cdot b_m| \le |b_n|\cdot |a_n - a_m| + |a_m|\cdot |b_n - b_m|$