I have a question given to me to practice for the exam but it's not a homework :)
Consider the following series:
$\sum_{n=1}^\infty (-1)^nln(1+ \frac{x}{n(1+x)})$, $x$ $\in [0, \infty)$
Study its pointwise and normal convergence, and show that it converges uniformly.
What our professor did for pointwise convergence is the following:
$f_n(x)$ = $(-1)^n\frac{x}{n(1+x)}$ $+$ $O(\frac{1}{n^2})$ ; ( he used taylor expansion of $ln(1+x)$ near $0$).
$\sum_{n=1}^\infty \frac{(-1)^n}{n}$ converges by the alternated series test ( since $\frac{1}{n}$ is decreasing and tending to zero), and $\sum_{n=1}^\infty \frac{1}{n^2}$ converges by the $p-test$. Hence $\sum_{n=1}^\infty f_n$ converges.
I can understand every step here but my question is that what is the name of this method? (as this is the first time for me to see it) and when can we use it? Is it the case when we have one of the functions of known taylor expansions near zero and we have $O(something)$ whose series converges? SORRY but my prof. sent the solution for this part without explaining in class.
According to normal convergence I know its two conditions:
1) $\exists$ $N$ such that $\forall$ $n > N, f_n$ is bounded.
2) $\sum_{k=0}^\infty \|f_k\|_\infty$ converges ( which's a numerical series).
But I can't do anything studying it :(
I think showing uniform convergence will depend on studying normal convergence first, since normal convergence $\Longrightarrow$ uniform convergence. So if we have the normal convergence of this series then we are done, but if we have no normal convergence we should check if $\|f_n\|_\infty$ tends to zero. If not, then we have no uniform convergence, but if it tends to zero we should check if $\|R_n\|_\infty$ tends to zero, if yes then we are done, if no then we have no uniform convergence. I think I know the ideas well but the problem lies in that the series is not easy as it contains $(-1)^n$.
Any help?
Normal convergence: Note
$$\sup_{x\in [0,\infty)} |f_n(x)| \ge \lim_{x\to \infty}|f_n(x)| = \ln (1+1/n).$$
Since $\sum \ln (1+1/n) = \infty,$ normal convergence fails on $[0,\infty).$
Uniform convergence: We have
$$f_n(x) = \frac{(-1)^nx}{n(1+x)} + R_n(x),$$ where $|R_n(x)|\le C/n^2,$ with the constant $C$ independent of $x\in [0,\infty).$ (You should verify the "independent of $x$" claim.) It follows that $\sum R_n(x)$ converges uniformly (in fact normally) on $[0,\infty).$ So we need only worry about $\sum \dfrac{(-1)^nx}{n(1+x)}.$ For each $x\ge 0$ this is an alternating series, with terms decreasing to $0$ in absolute value. In such a situation, the "first term dominates". So for any $N\in \mathbb N,$
$$\left |\sum_{n=N}^{\infty}\frac{(-1)^nx}{n(1+x)}\right |\le \left |\frac{(-1)^Nx}{N(1+x)}\right | = \frac{1}{N}\cdot \frac{x}{1+x} \le \frac{1}{N}.$$
This proves that our series is uniformly convergent on $[0,\infty).$