I'm having some troubles finishing off the last part of the following problem.
Let $\{u_n\}_{n\geq 1}$ be the sequence of solutions to the Cauchy problem $$\begin{cases} u'=-\log(u^2+1/n) \\ u(0)=1 \end{cases}$$ where $t\geq 0$. Discuss the convergence of the following series: $$\sum_{n=1}^{\infty}(-1)^n (1-u_n(t)).$$
Here's what I did. Notice that each term of the series is actually well defined for all $t\geq 0$, since each $u_n(t)$ is globally defined on $[0,+\infty)$. Indeed, this is a consequence of the fact that $$f(t,u)=\log(u^2+1/n)$$ is continuous and sublinear (maybe it's an overkill but I couldn't think of an easier way to prove global existence). Hence, the solution is unique for any initial condition. In particular, notice that $$u(t)\equiv \sqrt{1-\frac{1}{n}}$$ is a solution to the differential equation, so by uniqueness our solution $u_n(t)$ lies above this line as $u_n(0)=1$. Then this implies that $u_n(t)$ is decreasing, and it's easy to check that indeed $$\lim_{t\to +\infty} u_n(t)=\sqrt{1-\frac{1}{n}}.$$ Now we can start tackling the convergence part. Pointwise convergence shouldn't be an issue: unless I did something wrong, we should be able to prove pretty easily that $u_{n+1}(t) \geq u_n(t)$ for all $t\geq 0$, and clearly $$0\leq 1-u_n(t)<1-\sqrt{1-\frac{1}{n}}\to 0$$ thus the series converges for all $t\geq 0$ by Dirichlet's test for alternating series. However, I don't know how to approach the uniform/absolute convergence. Intuitively it can't happen on all $[0,+\infty)$ as $$1-\sqrt{1-\frac{1}{n}}\sim \frac{1}{2n}$$ but this leaves open the question on what happens to compact sets of the form $[0,a]$. By monotonicity the problem is in $u_n(a)$, however I can't find any good estimate of how fast the convergence actually is.