I have the following two sums:

\(\displaystyle u(x) = \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1)}\)

\(\displaystyle u_{n}(x) = \sum_{n=1}^{N} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1)}\)

It sounds trivial but how do I (sufficiently) argue that \(\displaystyle u_{n}(x) \to u(x)\)

I have a question about notation here. When one sees $u(x)$ that is $u$ is a function of $x$.

However in $u(x) = \sum\limits_{k = 1}^\infty {\dfrac{{{{( - 1)}^k}}}{k}{1_{[k,k + 1)}}}$ there is no $x$ on the RHS.

I assume that it

**should be** the indicator function: ${\large 1}_{[k,k+1)}(x)=\begin{cases} 1 &: x\in[k,k+1) \\0 &: \text{ else }\end{cases}$

Now that really makes things odd. Because$k$ is an index in the sum it is a positive integer.

Thus if $x> 0$ then there is exactly one integer $J = \left\lfloor x \right\rfloor < x < (\left\lfloor x \right\rfloor + 1) = J + 1$

That means $u(x)=\dfrac{(-1)^{\left\lfloor x \right\rfloor}}{\left\lfloor x \right\rfloor}$ for $x>0$ and zero otherwise.

Now consider the sequence $u_n(x)$. If $x<N$ then $u_n(x)=0$ but if $x\ge N$ then $u_n(x)=\dfrac{(-1)^{\left\lfloor x \right\rfloor}}{\left\lfloor x \right\rfloor}$

Perhaps I have grossly misread the function? But in any case I find the an odd question.