Recently I've been blocked on a problem as follow :

Let $a_n\in \mathbb{R} $ and $ a ≤ |a_n| ≤ b$ In this case $a \ne b$ so obviously $a<b$

(In the specific question both $a<b<1$ and I guess that if they are over $1$ the series diverges)

I don't know what I should do to prove it, I guess it has something to do with $-b\le a_n \le b $ or $0\le |a_n|\le b$ but I can't think of anything. The correct answer is that $|\sum_{n=0}^{\infty} a_n^n|$ converges. What's worse is that I should be able to find a value as $|\sum_{n=0}^{\infty} a_n^n|≤c$ with $c$ a reel number...

Edit: Here's the full statement: $1/4≤|a_n|≤1/2$ for every n ≥ 0 and the answer requested is : $∑a_n^n$ converge and $|∑a_n^n|≤2$ Thank you in advance.

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    $\begingroup$ welcome to Math Stack Exchange. Your statement is incorrect. For example: $$0\leq \left|\frac{1}{n}\right|\leq \frac{3}{4}\,\,\, \forall n\in \mathbb{N}$$ But:$$\sum_{n=1}^{+\infty}\frac{1}{n}$$ does not converge. $\endgroup$
    – Matteo
    Nov 25, 2021 at 19:32
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    $\begingroup$ What exactly is the statement of the problem? $\endgroup$ Nov 25, 2021 at 19:33
  • $\begingroup$ The statement is : $1/4 ≤ |a_n|≤ 1/2$ for every n ≥ 0 and the answer requested is : $\sum a_n^n$ converge and $ |\sum a_n^n| ≤ 2$ Yeah there is a problem in my question I forgot the power of n sorry $\endgroup$ Nov 25, 2021 at 19:37
  • $\begingroup$ @user15757055: this isn't included in your prompt. Rewiev it. Thanks. $\endgroup$
    – Matteo
    Nov 25, 2021 at 19:44
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    $\begingroup$ your statement isn't a question What is the question? Please update you answer with additional information even if you have posted them as comments. $\endgroup$
    – miracle173
    Nov 25, 2021 at 19:45


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