Uniform convergence of the following series:

Question

Does the following series converge uniformly ?$$\sum_{n=0}^{\infty} \frac{-1^n} {x+n} \ for \ (x\in R^+)$$ My thought:

for pointwise convergence: $$ \lim_{n \rightarrow \infty} \frac{-1^n} {x+n} =0 $$ so $$|f_n(x) - f(x)|=|\frac{-1^n} {x+n}|$$ now for the function $$g(x)=\frac{-1^n} {x+n} \implies g'(x)= \frac {-(-1)^n}{(x+n)^2}\neq0 $$

how can I proceed further ?

what I am trying to do here is to find the critical point to maximize the function $g(x)$ but clearly there is no critical point as the derivative can't be zero.

Answer 1

You are mistakenly trying to prove that the series converges uniformly by analyzing the convergence of the term sequence. For a series $\sum f_n(x)$ to converge uniformly it is necessary but not sufficient that $f_n(x) \to 0$ uniformly.

In this case, for all $x \in \mathbb{R}^+$ we have

$$|f_n(x)| =\left|\frac{(-1)^n}{x + n} \right| \leqslant \frac{1}{n}$$

and the convergence $f_n(x) \to 0$ is uniform. This only means we can't yet rule out uniform convergence of the series.

The series does, in fact, converge uniformly by Dirichlet's test since the partial sums $\sum_{n=0}^m (-1)^n$ are uniformly bounded and $1/(x+n)$ converges monotonically and uniformly to $0$ for $x \in \mathbb{R}^+$.

Answer 2

Note that $$ \begin{align} \sum_{k=2n}^\infty\frac{(-1)^k}{k+x} &=\sum_{k=n}^\infty\left(\frac1{2k+x}-\frac1{2k+1+x}\right)\\ &=\sum_{k=n}^\infty\frac1{(2k+x)(2k+1+x)}\\ &\le\sum_{k=n}^\infty\frac1{(2k-1)(2k+1)}\\ &=\frac1{4n-2} \end{align} $$ and $$ \begin{align} \sum_{k=2n+1}^\infty\frac{(-1)^k}{k+x} &=-\sum_{k=n}^\infty\left(\frac1{2k+1+x}-\frac1{2k+2+x}\right)\\ &=-\sum_{k=n}^\infty\frac1{(2k+1+x)(2k+2+x)}\\ &\ge-\sum_{k=n}^\infty\frac1{2k(2k+2)}\\ &=-\frac1{4n} \end{align} $$ Therefore, $$ \left|\,\sum_{k=n}^\infty\frac{(-1)^k}{k+x}\,\right|\le\frac1{2n-2} $$ independent of $x\ge0$.