# If a_n converges and lim_(n->oo) a_n -b_n=c, where c is a constant, does b_n converge?

May 10, 2018

${\lim}_{n \to \infty} {b}_{n} = - c + {\lim}_{n \to \infty} {a}_{n}$

#### Explanation:

Let:

${\lim}_{n \to \infty} {a}_{n} = L$

Then for any number $\epsilon > 0$ we can find ${N}_{\epsilon}$ such that:

$n > {N}_{\epsilon} \implies \left\mid {a}_{n} - L \right\mid < \frac{\epsilon}{2}$

Similarly, as ${\lim}_{n \to \infty} \left({a}_{n} - {b}_{n}\right) = c$ for the same $\epsilon$ we can find ${M}_{\epsilon}$ such that:

$n > {M}_{\epsilon} \implies \left\mid {a}_{n} - {b}_{n} - c \right\mid < \frac{\epsilon}{2}$

Consider now the quantity:

$\left\mid - {b}_{n} + L - c \right\mid = \left\mid {a}_{n} - {b}_{n} - c - {a}_{n} + L \right\mid$

using the triangular inequality:

$\left\mid - {b}_{n} + L - c \right\mid \le \left\mid {a}_{n} - {b}_{n} - c \right\mid + \left\mid - {a}_{n} + L \right\mid$

But then if we take ${P}_{\epsilon} = \max \left({M}_{\epsilon} , {N}_{\epsilon}\right)$ we have:

$n > {P}_{\epsilon} \implies \left\mid - {b}_{n} + L - c \right\mid < \frac{\epsilon}{2} + \frac{\epsilon}{2}$

which means:

${\lim}_{n \to \infty} {b}_{n} = L - c$