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$\sum_{n=1}^{\infty}x_n$ is a convergent series and $\sum_{n=1}^{\infty}y_n$ is a divergent series. Prove their sum diverges.

My attempt:

Suppose $\sum_{n=1}^{\infty}x_n + y_n$ converges.

Since $\sum_{n=1}^{\infty}-x_n = -\sum_{n=1}^{\infty}x_n$ converges, $\sum_{n=1}^{\infty}x_n + y_n - \sum_{n=1}^{\infty}x_n = \sum_{n=1}^{\infty}y_n$

This implies that $\sum_{n=1}^{\infty}y_n$ converges, which is a contradiction. Therefore $\sum_{n=1}^{\infty}x_n + y_n$ diverges.

How is this proof?

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  • $\begingroup$ You need parentheses and fewer displayed expressions. Otherwise fine. $\endgroup$
    – zhw.
    Mar 30, 2017 at 4:05
  • $\begingroup$ It looks perfectly acceptable. $\endgroup$
    – Mark Viola
    Mar 30, 2017 at 4:16
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    $\begingroup$ For typesetting, reserve the use of $$ expression $$ for things that you want to stand out by themselves in a line on its own in the center. For standard use, just use $ expression $. Also, keep the equals signs inside of math mode, no need to end and restart mathmode each time you come across an equals sign. $\endgroup$
    – JMoravitz
    Mar 30, 2017 at 6:15

2 Answers 2

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Yes, that would be the standard way of doing it.

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If $\displaystyle \sum_{n=1}^{\infty} x_{n}+y_{n}$ converges, then we can talk about the sequence of $c_{r}=\displaystyle \sum_{n=1}^{r} x_{n}+y_{n}$ in terms of its behavior for arbitrarily large $r$. We can rewrite it $c_{r}$ as $p_{r}+q_{r}$, where $p_{r}, q_{r}$ are the partial sum sequences of $x_{n}$ and $y_{n}$ respectively. This can be done because we are assuming that $r$ is finite and for any given $r$, they are identical.

Now, we know the behavior of $p_{r}$ in our desired range (somewhere between 0 and infinity but really big); it's $O(1)$. This is because we are given that $\sum_{n=1}^{\infty} x_{n}$ converges.

So for arbitrarily large $r$,

$c_{r}-q_{r}=O(1)$.

This means the behavior of $c_{r}$ is 'roughly' the behavior of $q_{r}$.

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