# Uniform convergence of the series $\sum_{n=1}^{\infty} \frac{1}{n^3}$

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.

• 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). Commented Dec 3, 2022 at 8:44
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.