$\sum_{n=1}^{\infty}x_n$ 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?

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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


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


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 $r$,


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.