TITLE: Prove that the sum of a convergent and a divergent sequence is divergent
QUESTION [5 upvotes]: I wonder whether my proof is correct. The exercise is to prove that for $(x_n)$ a convergent sequence and $(y_n)$ a divergent sequence, $(x_n + y_n)$ diverges.
Attempt (only for divergence to infinity):
The sequence $(x_n)$ is convergent, thus bounded, so for every $k \in \mathbb{N}$: $x_k \geq M$ for $M = \inf x_n$. Let $\xi > 0$. Choose $N \in \mathbb{N}$ such that $y_n \geq \xi - M$ for every $n \geq N$. Then: $x_n + y_n \geq M + y_n \geq M + \xi - M \geq \xi$.
In a same manner for divergence to infinity.
Is this proof correct? Thank you!

REPLY [4 votes]: Since $x_n$ is convergent we suppose it converges to $l$. Now assume $x_n+y_n$ also converges to $p$. Therefore according to definition:
$$\forall\epsilon>0\to \exists N\qquad \forall n>N\to |l-x_n|<\epsilon,|x_n+y_n-p|<\epsilon$$
and by using the triangle inequality we obtain:
$$|y_n-(p-l)|=|l-x_n+x_n+y_n-p|\le|l-x_n|+|x_n+y_n-p|<2\epsilon$$
which concludes that $y_n$ converges to $p-l$ and this is a contradiction. So our problem has been proved.