The limit of $ \sum_{n=1}^\infty (-1)^nf(\frac xn)$ at zero Let $f : [ 0,+\infty) \longrightarrow \Bbb{R}$ be a convex and decreasing function which is continuous at $0$ and $\lim_{x\longrightarrow +\infty}f(x) = -\infty$. Define $F(x) = \sum_{n=1}^\infty (-1)^nf(\frac xn)$. Is it true that we always have  $\lim_{x\longrightarrow 0^+} F(x) = \frac 12f(0)$. 
 A: $\sum_{n=1}^\infty (-1)^n \left[f(\frac xn)-f(0)\right]$ converges by the Leibniz criterion. By the same criterion, the limit of the series is contained between $f(0)-f(x)$ and $f(\frac x2)-f(x)$, giving the convergence to $0$ for $x\to0$.
Which kind of weak limit do you use? 

The Cesaro limit of a sequence is actually an often used tutorial exercise. For a sequence $(x_n)_{n\in\mathbb N}$ it is defined as
$$x_*=\lim_{N\to\infty}\frac{x_1+x_2+\dots+x_N}N.$$
If the sequence converges in the usual sense, then also $x_*=\lim_{n\to\infty} x_n$. If the sequence alternates between $0$ and $1$, then the Cesaro limit is $\frac12$.
The difference between the partial sums $\sum_{n=1}^N (-1)^n f(\frac xn)$ and $\sum_{n=1}^N (-1)^n \left[f(\frac xn)-f(0)\right]$ oscillates between $-f(0)$ for odd $N$ and $0$ for even $N$. Thus the Cesaro limit is the limit of the convergent sequence minus $\frac12f(0)$,
$$\lim_{N\to\infty}\sum_{n=1}^N\left(1-\tfrac{n-1}N\right)(-1)^n f\left(\tfrac xn\right)=-\tfrac12f(0)+\sum_{n=1}^\infty (-1)^n \left[f\left(\tfrac xn\right)-f(0)\right].$$
