# If $t<0$, what is $t\sum a_n$?

Let $$t$$ be a nonpositive real number (i.e. $$t<0$$) and $$\{a_n\}$$ be a nonnegative sequence if $$\sum a_n<\infty$$ then how do we prove or disprove that $$t\sum a_n<\infty?$$

• In this given case, $\sum a_n$ is just a real number. You could write $s=\sum a_n\in\mathbb{R}$ and since $t$ is some other real number, we have $ts=st\in\mathbb{R}$. (Note we're not considering $\infty$ to be a real number!)
– EBO
May 24, 2020 at 17:47

In your case, $$a_n \ge 0$$ and thus $$|a_n| = a_n$$. You trivially have that $$\sum_n |a_n| = \sum_n a_n < \infty .$$ Now, for negative $$t \in \mathbb{R}$$ you have $$t \sum_n a_n \le | t \sum_n a_n | \le |t| \left| \sum_n a_n \right| = |t| \sum_n |a_n | = |t| \sum_n a_n < \infty$$ (the middle equality is because that $$\forall n : a_n \ge 0$$). In a similar manner you obtain that $$-\infty < t \sum_n a_n$$ and thus you deduce that $$\sum_n a_n < \infty$$ converges, using the Direct comparison test.
Since $$a_n \geq 0$$ by assumption, we have: $$\infty > \sum a_n \geq 0$$ Now, if you multiply for some real number $$t < 0$$, the result will be: $$- \infty < t \sum a_n \leq 0 < + \infty$$ Notice that the sum is not infinite, because you have multiplied by a finite value $$t$$.