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 /> &lt; E.

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

1. i &gt; I, |S_i - A| &lt; E_1
2. j &gt; J, |Q_j - B| &lt; 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| &lt;= |S_n - A| + |Q_n - B| &lt; 2E_1

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

|S_n - A + Q_n -B| &lt; 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 /> &lt; E.

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

1. i &gt; I, |S_i - A| &lt; E_1
2. j &gt; J, |Q_j - B| &lt; 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| &lt;= |S_n - A| + |Q_n - B| &lt; 2E_1

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

|S_n - A + Q_n -B| &lt; 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!