Examine the uniform convergence of the series of functions $f_n: \mathbb{R} \to \mathbb{R}: x \mapsto \frac{(-1)^n}{n}$

Question

Examine the uniform convergence of the series of functions $f_n: \mathbb{R} \to \mathbb{R}: x \mapsto \frac{(-1)^n}{n}$

Attempt:

Note that for any $x \in \mathbb{R}$:

$$\sum_n^\infty f_n(x) = \sum_n^\infty \frac{(-1)^n}{n}$$

and by the alternating series criterium (Leibniz), the series $\sum_{n}^\infty f_n(x)$ converges for every $x \in X$. Hence, the series of functions $\sum_n^\infty f_n$ converges pointwise. Because the series does not depend on $x$, it also converges uniformly.

Is this correct?

Answer 1

Yes. The terms follow all the conditions of the Leibniz alternating series convergence theorem. The terms alternate in sign, decrease in magnitude, and tend to zero as $n$ approaches infinity. Since $(t_n = s_n - s_{n-1})$, we can say that $\sum_{n}^\infty f_n(x)$ does not depend on $x$.