Consider sequences $(a_n), (b_n), (c_n)$ in $\mathbb{R}$ with $\lim_n a_n=a<\infty, \lim_n b_n=b<\infty$ and $\lim_n c_n=\infty$.

1.) Show that $$ \lim(a_n+b_n-c_n)=\lim(a_n)+\lim(b_n)-\lim(c_n). $$

2.) Show the same statement with $\liminf_n$ instead.

How can I show that?

1.) I have to show that for any $\varepsilon>0$ it exists a $n_0\in\mathbb{N}: \lvert a_n+b_n-c_n-(a+b-\infty)\rvert\leq\varepsilon$.

  • $\begingroup$ Don't you mean sequences? $\endgroup$ – user112167 Nov 29 '13 at 21:45
  • $\begingroup$ Oh, yes, sequences! I correct it. $\endgroup$ – math12 Nov 29 '13 at 21:45
  • $\begingroup$ Try to show first that $\lim(a_n+b_n) = \lim(a_n)+\lim(b_n)$, then you will also know what $\lim(a_n+b_n-c_n)$ is. And remember that $(a+b-\infty)$ is not possible. $\endgroup$ – user112167 Nov 29 '13 at 21:50
  • $\begingroup$ I know how to proof $\lim(a_n+b_n)=\lim(a_n)+\lim(b_n)$. Thats very simple... $\endgroup$ – math12 Nov 29 '13 at 21:51
  • 1
    $\begingroup$ Do you also know what it means for a limit to tend to infinity in delta-epsilon terms? It is not possible because $a+\infty$ or $a-\infty$ is not defined. $\endgroup$ – user112167 Nov 29 '13 at 21:56

Let us give some hints. As the OP writes " know how to proof lim(an+bn)=lim(an)+lim(bn). Thats very simple... " we consider the sum


where $a_n+b_n:=\beta_n \rightarrow \beta=a+b$, for $n\rightarrow +\infty$, with $\beta$ finite. It follows that $\beta_n$ is bounded. We want to prove that

$$\beta_n-c_n\rightarrow -\infty$$

for $n\rightarrow +\infty$. In other words, we have to show that

$$ \forall \epsilon>0~ \exists N=N(\epsilon)~s.t.~\forall n\geq N \Rightarrow\beta_n-c_n<-\epsilon.$$

The prof is easily found on any book of Analysis. Alternatively, you need to use the fact that $\beta_n$ is bounded (in our case we need that $\beta_n\leq Q$, with $Q$ finite for all $n$) and $c_n\rightarrow -\infty$, for $n\rightarrow +\infty$.


Judging from the way the question (and the approach for solution by OP) has been formulated I see that the fundamental difficulty comes by treating the sequence $c_{n}$ which diverges to $\infty$ at par with sequence $a_{n}, b_{n}$ which converge to $a, b$ respectively. The question should be reworded that if sequences $a_{n}, b_{n}$ converge to some values $a, b$ respectively and sequence $c_{n}$ diverges to $\infty$ then show that the sequence $(a_{n} + b_{n} - c_{n})$ diverges to $-\infty$.

Since $a_{n}, b_{n}$ converge to $a, b$ we have a positive integer $m_{1}$ such that $|a_{n} - a| < 1, |b_{n} - b| < 1$ whenever $n > m_{1}$. This means that $a - 1 < a_{n} < a + 1, b - 1 < b_{n} < b + 1$ whenever $n > m_{1}$. And we see that $a_{n} + b_{n} < a + b + 2$ whenever $n > m_{1}$. Now let $N$ be any positive number so that $-N$ is negative. Since $c_{n}$ diverges to $\infty$ it means that there is a positive integer $m_{2}$ such that $c_{n} > N + a + b + 2$ whenever $n > m_{2}$. In other words $-c_{n} < -(N + a + b + 2)$ whenever $n > m_{2}$.

If $m = \max (m_{1}, m_{2})$ then $a_{n} + b_{n} - c_{n} < a + b + 2 - (N + a + b + 2) = -N$ whenever $n > m$. Thus we have shown that given any negative number $-N$ we are able to find a positive integer $m$ such that $(a_{n} + b_{n} - c_{n}) < -N$ whenever $n > m$. This means that $(a_{n} + b_{n} - c_{n})$ diverges to $-\infty$ as $n \to \infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.