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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 $$t\sum a_n<\infty?$$

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    $\begingroup$ 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!) $\endgroup$
    – EBO
    May 24, 2020 at 17:47

2 Answers 2

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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.

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  • $\begingroup$ I like the details. Thanks $\endgroup$
    – user27573
    May 24, 2020 at 17:50
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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 $t$.

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  • $\begingroup$ Nice ans straight forward answer. Thanks $\endgroup$
    – user27573
    May 24, 2020 at 17:49

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