can someone please help me prove this statement,
two sequences (an) and (bn) converge if and only if (an + bn) and (an - bn) converge.
thank you.
Say $\displaystyle \left( {a_n } \right) \to A\quad \& \quad \left( {b_n } \right) \to B$.
If $\displaystyle \varepsilon > 0$ find an $\displaystyle N$ such that $\displaystyle n \ge N\quad \Rightarrow \quad \left| {\left( {a_n } \right) - A} \right| < \frac{\varepsilon }{2}\quad \& \quad \left| {\left( {b_n } \right) - B} \right| < \frac{\varepsilon }{2}$.
You should understand how to do these if you are expected to complete these problems.
ok so how do you know that abs(an - A) + abs(bn - B) is less than epsilon? you know what i'm saying? we know that that expression and that epsilon are both greater than or equal to abs((an + bn) - (A+B)). how do we know that abs(an - A) + abs(bn - B) is less than epsilon?
you said that abs(an - A) is less than epsilon/2, but we don't know this, right?
it's like the triangle inequality doesn't work because it's reversed.
No I did not miss your point. But rather tried to help see the error of your ways.
Consider: $\displaystyle \left( {a_n } \right) = \left( { - 1} \right)^n \quad \& \quad \left( {b_n } \right) = \left( { - 1} \right)^{n + 1} $.
Is it true that $\displaystyle \left( {a_n + b_n } \right) \to \left( {1 - 1} \right)$?
Is either of these true: $\displaystyle \left( {a_n } \right) \to \left( 1 \right)\quad \mbox{or}\quad \left( {b_n } \right) \to \left( { - 1} \right)$.
In fact is it possible for either $\displaystyle \left( {a_n } \right)\quad \mbox{or}\quad \left( {b_n } \right)$ to converge.