series of two sequences Given $$ \sum_{n=1}^\infty(a_n+b_n)$$
converges, and $$a_n \to 0$$
prove that $$a_1 + b_1 + a_2 + b_2 + \cdots$$
converges.
The question's intention is to show that $a_n \to 0$ is a sufficient condition for the series to converge. 
My try:
using the fact that  $ \sum_{n=1}^\infty(a_n+b_n)$ converges, we conclude that $$a_n + b_n \to 0$$
by Cauchy theorem of series convergence.
applying limit arithmetic we conclude that $$b_n \to 0$$
for a given $\epsilon >0$, there exist $N>0$ such that for every $n>N$:
$$|a_n|<\epsilon ~~,~~ |b_n|<\epsilon$$ 
So trying to show that $$|a_{n+1}+b_{n+1} + \cdots + a_{n+p} + b_{n+p}|< \epsilon_0$$ will do the trick (Cauchy).
With the triangle inequality:
$$|a_{n+1}+b_{n+1} + \cdots + a_{n+p} + b_{n+p}| \le |a_{n+1}|+|b_{n+1}| + \cdots + |a_{n+p}| + |b_{n+p}| \lt 2 p \epsilon$$
and here i am stuck...
Note:
The partial sum $$S_n = \sum_{k=1}^n c_n$$ where $$c_{2n-1} = a_n~~,~~c_{2n}=b_n$$
may not converge. example:
$$a_n = (-1)^n ~~,~~ b_n=(-1)^{n+1}$$
without the parentheses the partial sum doesn't converge. 
so $$\sum_{n=1}^\infty (a_n + b_n) = (a_1 + b_1) + (a_2 + b_2) + \cdots \ne a_1 + b_1 + a_2 + b_2 + \cdots$$
 A: Let $(T_n)$ be the sequence of partial sums for $\displaystyle\sum_{n=1}^{\infty}(a_n+b_n),\;\;\;$ let $\displaystyle\lim_{n\to\infty}T_n=T,\;\;$and 
let 
$(S_n)$ be the sequence of partial sums for the series $a_1+b_1+a_2+b_2+a_3+b_3+\cdots$.
Then $S_{2n}=T_n$ for each n, so $\displaystyle\lim_{n\to\infty}S_{2n}=T$ and
$\displaystyle\lim_{n\to\infty}S_{2n+1}=\lim_{n\to\infty}(S_{2n}+a_{n+1})=T+0=T$ also.
Therefore $\displaystyle\lim_{n\to\infty}S_{n}=T$, so the series converges. 
A: Given $\varepsilon>0$, choose $M$ so large that if $m\ge M$ then $|a_m|<\varepsilon/2$ (possible because $a_n\to0$ as $n\to\infty$) and choose $N$ so large that if $n\ge N$ then
$$
|(a_1+b_1)+\cdots+(a_n+b_n)-\text{sum}|<\frac\varepsilon2. \tag 1
$$
If $n\ge\max\{M,N\}$ then
\begin{align}
& |(a_1+b_1)+\cdots+(a_n+b_n)+a_{n+1}-\text{sum}| \tag 2 \\[6pt]
\le {} & |(a_1+b_1)+\cdots+(a_n+b_n)-\text{sum}| + |a_{n+1}| < \frac\varepsilon2+\frac\varepsilon2 = \varepsilon.
\end{align}
Every partial sum of the series whose convergence is to be proved is of one of the forms $(1)$ and $(2)$, so all differ from the sum by less than $\varepsilon$ when enough terms are taken.
