Basic sequence convergence proof check Let a,b, $c \in \mathbb R$. Suppose that $\{a_n\}$ converges to a and that $\{b_n\}$ converges to b. By definition prove that the sequence $\{\pi a_n - 9cb_n\}$ converges.
My solution: \
CASE 1:
If c = 0, then $\{ c9b_n \} = \{0,0,\cdots, 0\} \Rightarrow 0 = 09b$. But this still converges for $\{ \pi a_n \} \Rightarrow \pi a$
$\exists a \in \mathbb R, \forall \epsilon > 0, \exists N > 0$ such that for all $n \in \mathbb N$, if $n > N$, then $|a_n - a| < \frac{\epsilon}{\pi}$
Suppose $n > N$, then
$|\pi a_n - \pi a| = |\pi (a_n - a)| < \pi \frac{\epsilon}{ \pi} = \epsilon$
Case 2:
Assuming $c \neq 0$
(1) $\exists a \in \mathbb R, \forall \epsilon > 0, \exists N_1 > 0$, such that for all $n \in \mathbb N$, if $n > N_1$, then $|a_n - a| < \frac{\epsilon}{2\pi}$
(2) $\exists b \in \mathbb R, \forall \epsilon > 0, \exists N_2 > 0$, such that for all $n \in \mathbb N$, if $n > N_2$, then $|b_n - b| < \frac{\epsilon}{18|c|}$
Let $\epsilon > 0$ be arbitrary
Choose N = max$(N_1, N_2) > 0$
Suppose $n > N$, then 
$$|\pi a_n -c9b_n - (\pi a - 9bc)| 
$$= \pi \frac{\epsilon}{2 \pi } + 9|c| \frac{\epsilon}{18 |c|}$$
$$= \epsilon/2 + \epsilon/2 = \epsilon$$ As required
Never worked with this type of proof before. Could someone verify, thx.
 A: Very well constructed proof, with some minor glitches (I could just edit them out, but it's good that you see them:


*

*What about if $c=0$? Then $\frac{\epsilon}{18|c|}$ is not a well defined term...

*In the second row, instead of $|\pi(a_n)-a|$, you should have $|\pi(a_n-a)|$

*In the second to last row, you should instead of $9c\frac{\epsilon}{18|c|}$ have $9|c|\frac{\epsilon}{18|c|}$


Other than that, yes, your proof is very good. It clearly shows that the sequence converges to $\pi a- 9cb$

Edit:
Your proof is now complete. The only thing left is you have to move the "Let $\epsilon > 0$ be arbitrary" to the beginning of the proof.
Just for an alternative (slightly shorter) version, you could also tackle the $c=0$ situation in this way:
Let $\epsilon > 0$ be arbitrary
(1) $\exists a \in \mathbb R, \forall \epsilon > 0, \exists N_1 > 0$, such that for all $n \in \mathbb N$, if $n > N_1$, then $|a_n - a| < \frac{\epsilon}{2\pi}$
(2) If $c\neq 0$, then $\exists b \in \mathbb R, \forall \epsilon > 0, \exists N_2 > 0$, such that for all $n \in \mathbb N$, if $n > N_2$, then $|b_n - b| < \frac{\epsilon}{18|c|}$
Let $\epsilon > 0$ be arbitrary
Choose $N = \max(N_1, N_2) > 0$
Suppose $n > N$, then 
$$|\pi a_n -c9b_n - (\pi a - 9bc)| = |\pi (a_n - a) + 9c(b - b_n)|$$
$$\leq |\pi (a_n - a)| + |9c(b_n - b)| \text{ by triangle inequality}$$


*

*We know that $|\pi (a_n - a)| \leq \pi \frac{\epsilon}{2 \pi }\leq \frac\epsilon2$

*if $c=0$, then $|9c(b_n-b)| = 0\leq \frac{\epsilon}{2}$, and if $c\neq 0 $, then $|9c(b_n-b)|\leq 9|c| \frac{\epsilon}{18 |c|}=\frac\epsilon2$. In both cases, $|9c(b_n-b)|\leq \frac\epsilon2$.


Therefore, $$|\pi (a_n - a)| + |9c(b_n - b)|\leq \epsilon/2 + \epsilon/2 = \epsilon$$ as required
