proof that $\sum z_n$, $\sum w_n$ convergent, then $\sum (z_n+w_n)$ converges

I need to prove that if

$\sum z_n$, $\sum w_n$ are convergent, then $\sum (z_n+w_n)$ converges.

What I did:

$$\sum z_n = \lim_{k\to \infty}\sum_{n=0}^kz_n\\\sum w_n = \lim_{k\to \infty}\sum_{n=0}^kw_n$$

Shouldn't I just say that:

$$\lim_{k\to \infty}\sum_{n=0}^k(z_n+w_n)\mbox{ converges }\iff \lim_{k\to \infty}\sum_{n=0}^kz_n \mbox{ and } \lim_{k\to \infty}\sum_{n=0}^kw_n \mbox{ both converge?}$$

I'd to the same to prove that if $\sum_{n=0}^{\infty} w_n$ converges, then $\sum_{n=0}^{\infty} c\cdot w_n$

• You'll need to check limits of partial sums. – Sean Roberson Nov 9 '16 at 21:14
• The iff part doesn't hold. Let $z_n=n$ and $w_n=-n$. – Alex R. Nov 9 '16 at 21:14
• Note that the implication only works in one direction. That's because you would be able to say that if $\sum a_n$ converges, then so do $\sum (a_n-\textrm{anything}_n)$ and $\sum\textrm{anything}_n$ which is obviously false. – MPW Nov 9 '16 at 21:19

If we know that a series converges that means that the sequence of partial sums go to zero. (Note if the partial sums go to 0 this doesn't imply convergence. Take $\sum \frac{1}{j}$ for instance). In this case, we know that the sequences $z_n$ and $w_n$ go to zero (because they are given as converging) and we know that the sum of two converging sequences also converges, thus $\sum (z_n+w_n)$ converges.
Hint: Suppose $\sum z_n=S$ and $\sum w_n = T$. This means that $$\left|\sum^N z_n-S\right|$$ can be made as small as desired by taking $N$ sufficiently large, and that $$\left|\sum^N w_n-T\right|$$ can be made as small as desired by taking $N$ sufficiently large.
You want to show that $\sum(z_n+w_n)$ converges to something, call it $U$. This means you want to show that $$\left|\sum^N (z_n+w_n)-U\right|$$ can be made as small as desired by taking $N$ sufficiently large. An obvious candidate is to try using $U=S+T$, and to note that since the sum is finite, it can be rearranged to get $$\left|\sum^N (z_n+w_n)-U\right| = \left|\left(\sum^N z_n -S\right)+\left(\sum^N w_n - T\right)\right|$$ The triangle inequality (which says $|a+b|\leq|a|+|b|$) then gives $$\left|\sum^N (z_n+w_n)-U\right| \leq \left|\left(\sum^N z_n -S\right)\right|+\left|\left(\sum^N w_n - T\right)\right|$$ You already know how to make each of these sums as small as desired individually. Can you furnish the remaining details?
As I mentioned in a comment, note that the reverse implication isn't true. However, your problem isn't to show "$\iff$", only "$\Longrightarrow$".