Answer to Question #111295 in Real Analysis for Deha

Question

Show that if X and Y are sequences such that X and X + Y are convergent, then Y is convergent.

Answer 1

Since X = {x_n} n=1,2,.. is convergent< we obtain that there exists A = $\lim\limits_{n\to\infin} x_n$. So, for every $\varepsilon$>0 there exists such N₀, that for every n>N₀ the following is true: |x_n - A|<$\frac{\varepsilon}{2}$ . (1)

We obtain the same for X+Y = {x_n + y_n}, n = 1,2,.. and B = $\lim\limits_{n\to\infin} (x_n + y_n)$ :

for every $\varepsilon > 0$ there exists such N₁, that for every n>N₁ the following is true:

|x_n + y_n - B|<$\frac{\varepsilon}{2} (2)$

Let us now denote with N = max{N₀, N₁}. Let's show that Y is convergent with the following limit:

B-A = $\lim\limits_{n\to\infin} y_n$ . For this, let's check the definition:

We will prove that for every $\varepsilon > 0$ there exists such N₂, that for every n>N₂ the following is true:

|y_n - (B-A)|<$\varepsilon$ (the definition of the limit): (assuming n>N (1) and (2) are both true)

|y_n - (B-A)| = |(x_n - A) - (x_n + y_n - B)| $\leqslant$ |x_n - A| + |x_n + y_n - B| $\leqslant$ $\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$. So for n>N this is true.

Exactly what we needed to show.