$\sum_{n=1}^{\infty}x_n$ 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?
$\endgroup$ 32 Answers
$\begingroup$Yes, that would be the standard way of doing it.
$\endgroup$ $\begingroup$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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