# Show $\{ a_n-b_n\} \rightarrow 0$ then $\{b_n\} \rightarrow L$

Suppose ${a_n}$ and ${b_n}$ are sequences with $\{a_n\}\rightarrow L$. Show $\{ a_n-b_n\} \rightarrow 0$ then $\{b_n\} \rightarrow L$

Not sure if my proof is sufficient.

Proof

Since $\lim{a_n}=L, \exists N_1$ for $n>N_1$

$|a_n - L| < \epsilon$

Now suppose $\lim{a_n - b_n} = 0, \exists N_2$ for $n>N_2$

$|(a_n - b_n) - 0| < \epsilon \Rightarrow |a_n-b_n|<\epsilon$

$\Rightarrow a_n - b_n < \epsilon \Rightarrow a_n<\epsilon + b_n \Rightarrow a_n-L < \epsilon + b_n - L$ Thus $|a_n - L| < \epsilon + b_n - L$

Let $M = max\{N_1,N_2\}$ and $n>M$ Then,

$|a_n-L+L-b_n| \leq |a_n-L| - |b_n-L| < \epsilon + b_n - L -|b_n - L| = \epsilon$

• What is $b$, here? – Bernard Oct 7 '15 at 11:41
• Just because $a_{n} - L < \epsilon + b_{n} - L$, you can't say then that $|a_{n} - L| < \epsilon + b_{n} - L$, because $a_{n} - L$ could be a negative number that's larger in magnitude than $\epsilon + b_{n} - L$. For example, $-2 < 1$, but $|-2| \not < 1$, since $|-2| = 2$. – layman Oct 7 '15 at 11:49
• Instead, you should have said $|a_{n} - b_{n}|< \epsilon \implies -\epsilon < a_{n} - b_{n} < \epsilon \implies -\epsilon + b_{n} < a_{n} < \epsilon + b_{n} \implies$ $$-\epsilon + b_{n} - L < a_{n} - L < \epsilon + b_{n} - L,$$ but then where could you go from here? – layman Oct 7 '15 at 11:50
• You need to choose $N_1$ so that $|a_n-L| < \epsilon/2$ for $n > N_1$, and $N_2$ so that $|a_n-b_n| < \epsilon/2$ for $n > N_2$. Then you have $|b_n-L| \le |a_n-L| + |a_n-b_n| < \epsilon$ for $n > \max(N_1,N_2)$. – TonyK Oct 7 '15 at 11:53

In fact, you need to prove that $\vert b_n - L \vert \to 0$. There are issues in your proof. In particular what are you using in order to deduce the inequality? $$|a_n-L+L-b_n| \leq |a_n-L| - |b_n-L|$$ which is wrong in general, i.e. if $a_n-L=1=L-b_n$.
You should try to use: $$\vert b_n-L \vert=\vert (b_n-a_n)+(a_n-L) \vert \le \vert a_n-b_n \vert + \vert a_n-L \vert$$ and then apply the good usage of $N_1, N_2$ as in your OP.