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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 $n_0\in\mathbb{N}: \lvert a_n+b_n-c_n-(a+b-\infty)\rvert\leq\varepsilon$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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

$$\beta_n-c_n$$

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 $n\rightarrow +\infty$.

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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 $n \to \infty$.

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