Uniform and absolute convergence of the given series

Question

Determine the convergence (uniform, absolute, conditional) of the series $\sum f_n(x)$, where $$f_{n}(x) = (-1)^n \frac{1}{x+n}~~~ (x \geq 0).$$

My Approach: Letting $g_n(x) = 1/(x+n)$, we see that $$g_n(x) - g_{n+1}(x) = \frac{1}{(x+n+1)(x+n} > 0~.$$ So $\{ g_n \}$ form a decreasing sequence. Clearly $\lim_{n \rightarrow \infty} g_n(x)= 0$ for a fixed $x$. Hence, by the alternating series test the given series converges.

Would someone help me in determining the uniform and absolute convergence of the series. Thanks in advance.

Answer 1

• Regarding absolute convergence: try lower bounding $\sum_n \frac{1}{x+n}$ with some tail of the harmonic series $\sum_m \frac{1}{m}$
• Regarding uniform convergence: the worst-case error of the partial sum $\sum_{n=1}^{N-1} f_n(x)$ over all $x$ is $\sup_{x \ge 0}\left| \sum_{n \ge N} f_n(x)\right| \le \sup_{x \ge 0} |f_N(x)| = \frac{1}{N}$

Answer 2

The series does not converge absolutelyy, since $\left\lvert\frac{(-1)^n}{x+n}\right\rvert=\frac1{x+n}$ and$$\lim_{n\to\infty}\frac{\frac1{x+n}}{\frac1n}=1$$and the harmonic series diverges.

And it converges uniformly, by Dirichlet's test:

• the sequence $\left(\sum_{n\leqslant N}(-1)^n\right)_{N\in\mathbb N}$ is uniformly bounded;
• the sequence $\left(\frac1{x+n}\right)_{n\in\mathbb N}$ is monotonic for each $x\geqslant0$;
• the sequence $\left(\frac1{x+n}\right)_{n\in\mathbb N}$ converges uniformly to the null function in $[0,\infty)$.