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These are homework questions I was assigned to do:

  1. Let $(x_{n})_{n \geq 0} $ be a convergent sequence in $\mathbb{R}$ with limit $x$ and let $a,b \in \mathbb{R}$ with $a \leq b$. Show that if $a \leq x_{n} \leq b$ for all $n$, then it is also true that $a \leq x \leq b$.

    I tried proving this by means of the triangle inequality (both the "regular" theorem and its reverse) and I tried finding a proof by contradiction, but to no avail. Could you help me out?

  2. Let $(x_{n})_{n \geq 0}$ be a convergent sequence and $(y_{n})_{n \geq 0}$ be a divergent sequence. Show that $(x_{n} + y_{n})_{n \geq 0}$ diverges.

    I tried proving this with the definition of both convergent and divergent sequences, assuming that $(x_{n} + y_{n})_{n \geq 0}$ converges and then I tried to find a contradiction, but I couldn't find one.

Thanks in advance,

Max Muller

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up vote 2 down vote accepted

Contradiction seems like the way to go. Suppose $x_n = c = b + \gamma > b$. The limit exists, so there is a $\epsilon$ ball for some $N$ so that all the terms of the sequence are in it. But if $\epsilon < \gamma/2$, say, we have a contradiction, as no $x_n > b$.

Do the same but on the other side for minimum.

Alternatively, if you know limsup and liminf, replicate the proof that the limit lies between them.

For the second one, you can go straight from the definition. Since the convergent converges to $L$, say, consider $N$ so that the divergent sequence gets bigger than $M - |L|$. It diverges, each of these exist, and so it still diverges.

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thank you! could you also please take a look at the second question I added later? – Max Muller Nov 20 '11 at 15:31

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