Q. 68 - Calculus

Question

Let $\left\{a_k\right\}$ and $\left\{b_k\right\}$ be convergent sequences with $a_k\to L$ and $b_k\to M$ as $k\to \infty$ and let $c$ be a constant. Prove the indicated basic limit rules from Theorem 7.11. You may wish to model your proofs on the proofs of the analogous statements from Section 1.5.

Prove that $\left(a_k+b_k\right)\to L+M$.

Step-by-Step Solution

Answer

Hence, the theorem is proved.

Step 1. Given Information.

The objective is to prove that $a_k+b_k\to L+M$.

Step 2. Forming the equations.

We use the definition of convergence for the sequence $\left\{a_k\right\}$ and $\left\{b_k\right\}$.

The sequence $\left\{a_k\right\}$ converges to $L$.

For given $\epsilon >0$, there exists a positive integer $N$ such that

$\left|a_k-L\right|<\frac{\epsilon }{2}$ for $k\ge N.............\left(1\right)$

The sequence $\left\{b_k\right\}$ converges to $M$.

For given $\epsilon >0$, there exists a positive integer $P$ such that

$\left|b_k-M\right|<\frac{\epsilon }{2}$ for  $k\ge P..............\left(2\right)$

Step 3. Proving the theorem.

Choose a positive integer $A$such that $A=max(N,P)$.

$$\begin{gathered} \left|a_k+b_k-(L+M)\right|=\left|(a_k-L)+(b_k-M)\right| \\ \le \left|\left(a_k-L\right)\right|+\left|\left(b_k-M\right)\right| \\ <\frac{\epsilon }{2}+\frac{\epsilon }{2} for k\ge P \\ =\epsilon \end{gathered}$$

Thus, for$k\ge P,\left|a_k+b_k-(L+M)\right|<\epsilon$ and hence,  $\left\{a_k+b_k\right\}$is convergent.

Therefore, the value is $\underset{k\to \infty }{\lim }\left\{a_k+b_k\right\}=L+M$.