Question

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

Answer 1

`lim_(n->oo) b_n = -c+ lim_(n->oo) a_n`

Explanation:

Let:

`lim_(n->oo) a_n = L`

Then for any number `epsilon > 0` we can find `N_epsilon` such that:

`n > N_epsilon => abs (a_n-L) < epsilon/2`

Similarly, as `lim_(n->oo) (a_n-b_n) = c` for the same `epsilon` we can find `M_epsilon` such that:

`n > M_epsilon => abs(a_n-b_n -c) < epsilon/2`

Consider now the quantity:

`abs (-b_n +L - c) = abs(a_n-b_n -c-a_n+L)`

using the triangular inequality:

`abs (-b_n +L - c) <= abs(a_n-b_n -c)+abs(-a_n+L)`

But then if we take `P_epsilon = max(M_epsilon,N_epsilon)` we have:

`n > P_epsilon => abs (-b_n +L - c) < epsilon/2+epsilon/2`

which means:

`lim_(n->oo) b_n = L-c`