Sequence Convergence proof check I am trying to solve the following sequence question:


  Show that $\{a_n\}_{n=1}^\infty$ converges to $A$ if and only if $\{a_n - A\}_{n=1}^\infty$ converges to $0$.


Proof: Assume $\{a_n\}_{n=1}^\infty$ converges to $A$. Then for each epsilon > 0, there exist N such that for all n>=N, abs($\{a_n\}$- $A$) &lt; epsilon. Now consider the sequence $\{a_n - A\}_{n=1}^\infty$. For this sequence to converge, it must be that for each epsilon > 0, there exist P such that for each n>=P, abs($\{a_n - A\}$ - $B$) &lt; epsilon. Let $B$ = 0, so we have abs($\{a_n\}$- $A$) &lt; epsilon. Since we know that abs($\{a_n\}$- $A$) &lt; epsilon when n>=N, it follows that N=P. Therefore, since there exist this P where for all n>=P, abs($\{a_n - A\}$ - $B$) &lt; epsilon, $\{a_n - A\}_{n=1}^\infty$ converges to $0$.

This is only the proof for the first assumption.
(1) $a_n \to A \Rightarrow \forall \epsilon&gt;0, \exists N$ such that $|a_n-A|&lt; \epsilon, \forall n &gt;n$. To have $(a_n-A) \to 0$ then we need that for a given $\epsilon, \exists N_0$ s.t $|a_n - A - 0| = |a_n - A|&lt; \epsilon$ for all $n &gt;N_0$. Just take $N_0 = N$ from above. 

(2) If $(a_n-A) \to 0$ then given $\epsilon&gt;0$ there exists $N_1$ s.t $|a_n - A - 0| = |a_n-A|&lt; \epsilon$ for all $n&gt;N_1$ but again if you choose $N = N_1$ in the definition of $a_n \to A$ then you are done.
Looks good. You can actually prove both directions at once as follows:

$\{a_n\}_{n=1}^{\infty}$ converges to $A$ if and only if for every $\epsilon &gt; 0$, there exists $N \in \mathbb{N}$ such that $\vert a_n - A \vert &lt; \epsilon$, which is true if and only $\{a_n - A\}_{n=1}^{\infty}$ converges to $0$.