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.
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.