I'm trying to find a rigorous and right way to prove my idea. If anyone could tell me how to do so that it's not too long but also makes sure that everything written is true, I would really appreciate it!


Let $\sum_{n \geq 1} a_n$ be a positive and convergent sequence, and let $(b_n)$ be a sequence such that $$b_n = 0 \text{ or } b_n = a_n , \forall n \geq 1$$

Show that $\sum_{n \geq 1} b_n$ converges.

Solution (attempt)

Let $\sum_{n \geq 1} a_n$ be a positive and convergent serie, and let $(b_n)$ be a sequence such that $$b_n = 0 \text{ or } b_n = a_n , \forall n \geq 1$$

Then, by definition of positive serie, we have that

$$a_n \geq 0, \forall n \in \mathbb N$$

And by definition of $b_n$,

$$b_n = 0 \leq a_n$$ $$\text{ or }$$ $$b_n = a_n \geq 0$$ $$\forall n \in \mathbb N$$

Therefore, we can say that {$b_n: n \in N$} is bounded above by $a_n$ and bounded below by $0$.

So, $0 \leq b_n \leq a_n$, $\forall n \geq 1$ which implies that $b_n$ is a positive sequence, and his associated serie is a positive serie.

By the comparaison test, if $0 \leq b_n \leq a_n$, $\forall n \geq 1$ and $\sum_{n \geq 1} a_n$ converges, then $\sum_{n \geq 1} b_n$ converges $\square$


I feel like I wrote way too much for this proof and the fact that $0 \leq b_n \leq a_n$, $\forall n \geq 1$ is not proven in a very convincing way. Is there a more rigorous way to prove this, another line of idea or maybe a better way of showing that $0 \leq b_n \leq a_n$, $\forall n \geq 1$??

Thank you!

  • $\begingroup$ You wrote $\forall n\le 1$. isn't $\ge$. $\endgroup$ – hamam_Abdallah Oct 27 '18 at 19:48
  • $\begingroup$ yes Ill correct that right away thank you $\endgroup$ – Ian Leclaire Oct 27 '18 at 19:50
  • $\begingroup$ If $b_n=a_n$, then $b_n\geq0$, so $0\leq b_n\leq a_n$. If $b_n\not=a_n$, then $b_n=0\leq a_n$, so $0\leq b_n\leq a_n$. There are no other cases, therefore $0\leq b_n\leq a_n$. This is basically what you wrote. It uses the law of excluded middle. $\endgroup$ – Melody Oct 27 '18 at 20:11

You don't need to bound the terms when you can already bound the sums.

$$ 0 \le \displaystyle\sum_{i=1}^n b_n \le \displaystyle\sum_{i=1}^na_n$$. Since this is true for the partial sums, it will be true in the limit. Then, by the comparison test, $\displaystyle\sum b_n$ must converge.

  • 1
    $\begingroup$ He has to bound the terms below by $0$ since otherwise the series might only be conditionally convergent (in which case this type of argument will fail). $\endgroup$ – Clayton Oct 27 '18 at 22:42

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