Limiting variable in interval: Lebesgue Dominated Convergence

Question

So I am pretty comfortable using the LDCT for definite integrals and summations, but I am looking at a problem that has the interval as a function of the limiting variable, i.e.:

$$\lim_{n\to\infty} \sum_{k=1}^n \frac{\sin\Big(\pi\sqrt{\tfrac{k}{n}}\Big)}{\sqrt{kn}}$$

In general can/how do you use LDCT for problems of the form:

$$ \lim_{n\to\infty} \sum_{k=1}^n f_n(k) \\ \lim_{n\to\infty} \int_0^n f_n(x)dx $$

*Note it is possible this is not an LDCT problem, it was just my first impression of it. The latter general question still stands regardless.

Answer 1

If one defines the functions $\{g_n:n\in\mathbb N\}$ as $$g_n(x)=\begin{cases}f_n(x)&\text{if }0\leq x\leq n\\ 0&\text{otherwise,}\end{cases}$$ or $g_n=f_n\cdot\chi_{[0,n]}$, where $\chi_A$ denotes the indicator function of a set $A$, then it is clear that $$\lim_{n\to\infty}\int_0^n f_n(x)~dx=\lim_{n\to\infty}\int_{\mathbb R}g_n(x)~dx.$$ (Note that the measurability of $g_n=f_n\cdot\chi_A$ is not an issue as long as $A$ is measurable, which is the case with $[0,n]$)

At this point, all you need to do to apply the LDCT is to find a function $g$ that dominates $g_n$ for each $n$ (i.e. that dominates $f_n$ on the compact $[0,n]$ for each $n$). Of course, if you find a function $g$ that dominates the $f_n$ almost everywhere, $g$ also dominates the $g_n$.

Answer 2

Note that if you can find a function $g$ such that $|f_n| \leq g$ and $g$ is Lebesgue integrable, then you can pull the limit inside the integral. Now summation is also integration- w.r.t. the counting measure on $\mathbb{N}$. Hence similar tricks will apply.

Of course the tricks won't apply if you can't find such a $g$, but sometimes it's useful even in that situation: Assume that there exists such a $g$ and then pull the limit inside the integral/summation, and maybe you reach a contradiction. Thus LDCT helps quite a bit in these kinds of problems.

Another useful thing to keep in mind while solving problems involving summation is that maybe LDCT won't work, but Monotone convergence theorem works. Esp. if you are working with sums involving "positive terms" only.