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Could an increasing sequence have a finite limit? If yes, can you give an example with a solution?

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    it must not, an example is $$a_n=n^2$$ – 2017-01-07
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    @Dr.SonnhardGraubner One might say that your sequence *has* a limit, since it isn't displaying oscillatory behavior. It just doesn't have a *real* limit. – 2017-01-07
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    hm, that depends on your definition, one can say it has an improper limit – 2017-01-07
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    i think he means with limit a real number – 2017-01-07
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    @user46944 that is not how I see limits defined... – 2017-01-07
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    @Dr.SonnhardGraubner A sequence with a real number as a limit is called convergent. Having a limit means finite or infinite, that's the subtle difference between having a limit and being convergent. – 2017-01-07
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    yes i know this here at the university Leipzig we use that a Limit is a real number and $$-\infty$$ or $$\infty$$ are improper limits – 2017-01-07

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Consider this sequence:

$$a_n:=1-\frac1{2^n}$$

It is increasing and tends to $1$.

But not all increasing sequences have limits:

$$b_n:=n$$

This one diverges.

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    @user46944 :-) covered. – 2017-01-07
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    Correct me if I am wrong, but doesn't the second seqence have an infinite limit? – 2017-01-07
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    @N.S. not by my definition, which I imagine to be the general definition. – 2017-01-07
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    So if I understand right, you use the two notions "has limit" and "convergent" to mean the same thing...How do you say that a sequence has an infinite limit? – 2017-01-07
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    @N.S. I don't. I would say the limit diverges to infinity. But not that it exists. – 2017-01-07
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Yes, but you need have boundedness from above i.e. $a_{n}\leq C$ for some $C \geq 0$. Take the sequecne $a_{n}=1-1/n$ It is strictly increasing, but bounded from above by any $C\geq 1$ and has the limit 1. If you need some details:
As $\frac{1}{n} >\frac{1}{n+1}$ we have $ 1-\frac{1}{n}<1-\frac{1}{n+1} $ $\forall n \in \mathbb{N}$. Also, since $1/n \leq 1 $, we have $ |1-\frac{1}{n}|\leq 1+1/n \leq 2 $ $\forall n \in \mathbb{N} $. Since $a_{n}$ is increasing and bounded, there must exist a limit. We now guess that the limit is 1. we calculate: Let $\epsilon > 0 $ $$ |a_{n}-1|=|1-1/n+1|=1/n\leq1/N \leq\epsilon \,\, \forall n\geq N $$ if we choose $N \in \mathbb{N}$ such that $ 1/N< \epsilon$

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Increasing sequence can, but does not have to have a real limit. For example, $a_n=n$ does not have a real limit, whereas $a_n=1-\frac{1}{n}$ does have a limit, namely $1$.