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Hey I have a short question. The given series is $\sum_{n=1}^{\infty} \frac{1}{n^3}$. The question is: use Weierstrass M test to determine if the given series uniformly converge. I am a bit confused. This series does not depend on $z$. My question is can anyone help me solve this problem?

My attempt: Let say that $f\left(z\right) = \sum_{n=1}^{\infty} \frac{1}{n^3}$. No matter what is entered for $z$ in $f\left(z\right)$, the series remains constant. I know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges and we can say that $\frac{1}{n^3} \leq \frac{1}{n^2}$ for $n = 1, 2, 3, ...$. This holds for all $z$ in the complex plane. So the series $f\left(z\right) = \sum_{n=1}^{\infty} \frac{1}{n^3}$ is uniformly convergent everywhere in the complex plane.

Thanks in advance.

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    $\begingroup$ Yes, any convergent series of constants is uniformly convergent. $\endgroup$ Commented Dec 3, 2022 at 8:44
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    $\begingroup$ You are right to be confused, and I don't think your attempt is helpful, alas. I suspect that $\frac{z^n}{n^3}$ was intended. And are you sure you have quoted the question to us exactly? (I mean the statement of the question, not the symbols for the series). $\endgroup$ Commented Dec 3, 2022 at 8:44
  • $\begingroup$ Yes, this is the exact question (I have quoted the question exactly). $\endgroup$
    – Oliver4
    Commented Dec 3, 2022 at 8:52
  • $\begingroup$ " use Weierstrass M test to determine if the given series uniformly converge" is not english. $\endgroup$ Commented Dec 3, 2022 at 9:33
  • $\begingroup$ "Use the Weierstrass M-test to determine if the above converges uniformly." This is the exact sentence. $\endgroup$
    – Oliver4
    Commented Dec 3, 2022 at 9:36

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The question is probably some kind of mistake/typo. It really doesn't make sense to talk about Weierstrass M-test for this problem because all the terms are just constant functions.

Anyway, your attempt is correct: you have in fact found a convergent series $\sum_n \frac{1}{n^2}$ where each function in the original series is bounded by the corresponding term in your new series. So your attempt is a valid proof using the M-test even though the M-test isn't realistically helping in this case.

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