Question #111295
Show that if X and Y are sequences such that X and X + Y are convergent, then Y is convergent.
1
Expert's answer
2020-04-22T18:57:23-0400

Since X = {xn} n=1,2,.. is convergent< we obtain that there exists A = limnxn\lim\limits_{n\to\infin} x_n. So, for every ε\varepsilon>0 there exists such N0, that for every n>N0 the following is true: |xn - A|<ε2\frac{\varepsilon}{2} . (1)


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

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

|xn + yn - B|<ε2(2)\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 = limnyn\lim\limits_{n\to\infin} y_n . For this, let's check the definition:

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

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


|yn - (B-A)| = |(xn - A) - (xn + yn - B)| \leqslant |xn - A| + |xn + yn - B| \leqslant ε2+ε2=ε\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. So for n>N this is true.


Exactly what we needed to show.


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