Check my proof on limit of two sequences

[Samuelb88]

Homework Statement

Let $$S_n$$ and $$Q_n$$ be sequences and suppose $$\lim_{n\rightarrow +\infty} {S_n} = A$$ and $$\lim_{n\rightarrow +\infty} {Q_n} = B$$. Then $$\lim_{n\rightarrow +\infty} {(S_n + Q_n)} = A+B$$.

The Attempt at a Solution

*I am using "E" in place of ε.

Proof: I want to show for every E > 0 there is an N such that whenever n>N, $$|S_n + Q_n - (A+B)|<br /> < E$$.

Suppose $$\lim_{n\rightarrow +\infty} {S_n} = A$$ and $$\lim_{n\rightarrow +\infty} {Q_n} = B$$. Then for every $$E_1 > 0$$, there are numbers I,J such that whenever:

1. $$i > I, |S_i - A| < E_1$$
2. $$j > J, |Q_j - B| < E_1$$

Take N=max(I,J) to ensure both the inequalities 1. and 2. will hold. Then by the triangle inequality

$$|S_n - A + Q_n -B| <= |S_n - A| + |Q_n - B| < 2E_1$$

Let $$E_1 = \frac{1}{2} E$$. Then

$$|S_n - A + Q_n -B| < E$$ as required. Q.E.D.

Look good?

 

[Dick]

Homework Statement

Let $$S_n$$ and $$Q_n$$ be sequences and suppose $$\lim_{n\rightarrow +\infty} {S_n} = A$$ and $$\lim_{n\rightarrow +\infty} {Q_n} = B$$. Then $$\lim_{n\rightarrow +\infty} {(S_n + Q_n)} = A+B$$.

The Attempt at a Solution

*I am using "E" in place of ε.

Proof: I want to show for every E > 0 there is an N such that whenever n>N, $$|S_n + Q_n - (A+B)|<br /> < E$$.

Suppose $$\lim_{n\rightarrow +\infty} {S_n} = A$$ and $$\lim_{n\rightarrow +\infty} {Q_n} = B$$. Then for every $$E_1 > 0$$, there are numbers I,J such that whenever:

1. $$i > I, |S_i - A| < E_1$$
2. $$j > J, |Q_j - B| < E_1$$

Take N=max(I,J) to ensure both the inequalities 1. and 2. will hold. Then by the triangle inequality

$$|S_n - A + Q_n -B| <= |S_n - A| + |Q_n - B| < 2E_1$$

Let $$E_1 = \frac{1}{2} E$$. Then

$$|S_n - A + Q_n -B| < E$$ as required. Q.E.D.

Look good?

Look's just fine to me.

 

[╔(σ_σ)╝]

This is only true if A and B are finite (i.e not +/-infinity) . If they are then the proof is flawless :).

 

[Samuelb88]

This is only true if A and B are finite (i.e not +/-infinity) . If they are then the proof is flawless :).

Woops, forgot to mention that. :) Thanks, guys!