If $\sum x_{n}$ and $\sum y_{n}$ are convergent show that $\sum (x_{n}+y_{n})$ is convergent Please check my proof
because $\sum  x_{n}$ and  $\sum y_{n}$ are convergent 
then 
$\sum  x_{n}=|x_{1}+x_{2}.....|<\frac{\epsilon }{2}$ and
$\sum y_{n}=|y_{1}+y_{2}.....|<\frac{\epsilon }{2}$
then $\sum x_{n}+y_{n}$
$|x_{1}+x_{2}.....|+|y_{1}+y_{2}.....|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon $
 A: Here is maybe an atypical solution that I think is kind of neat and in some small way "the reason" why this is true.
View $f: \mathbb R \times \mathbb R \to \mathbb R$, defined by $f(x,y)=x+y$ as a continuous function. Let $a$ be the first limit and $b$ the second, with $a_n$ the partial sums of the first and $b_n$ the partial sums of the second. Then $\lim (a_n,b_n)$ converges $\iff$ $\lim a_n$ and $\lim b_n$ converge and $(a_n,b_n) \to (a,b)$.
Hence,
$$a+b=f(a,b)=f(\lim (a_n,b_n))=\lim f(a_n,b_n)=\lim_{n \to \infty} \left(\sum_{k=1}^{n}a_n+\sum_{k=1}^{n}b_n\right)=\lim_{n \to \infty} \sum_{k=1}^{n}(a_n+b_n).$$
remark: this would have worked for any continuous function. Say $f: \mathbb R  \times \mathbb R \to \mathbb R$ given by $(x,y) \mapsto xy$. Then $\sum_{n=1}^{\infty} a_n b_n$ converges when the two said sums do. This has very little to do with sums  (as evidenced by my bare-boned notation) and a lot to do with continuous functions and sequences.
Here is something more along your lines that you should work out:
the sequence of partial sums are both cauchy, so it will suffice to see that the desired sequence is cauchy: let  $n>k$.
$$\left|\sum_{i=1}^{n}(a_n+b_n)-\sum_{i=1}^{k}(a_n+b_n)\right|=\left|\sum_{k}^{n}(a_n+b_n) \right| \leq \left| \sum_{k}^{n} a_n+\sum_{k}^{n} b_n \right| \leq \left|\sum_{k}^{n}a_n \right|+\left|\sum_{k}^{n} b_n \right|.$$
Notice what happens to the right hand side as $k$ becomes sufficiently large.
