I'm trying to calculate a sequence of probabilities, using the Law of Total Probability. It looks something like this:
$$ P(X_n = x) = \sum_{k=1}^{\sqrt{n}} a_k^{(n)} \left(1+O\left(\frac{1}{n} \right)\right).$$
Each of the summands contains a relative error of $O(1/n)$.
Question: how to show that the sum of the error terms, $\sum_{k=1}^{\sqrt{n}} a_k^{(n)} O\left(\frac{1}{n} \right)$, is negligible ? Clearly, $$\sum_{k=1}^{\sqrt{n}} a_k^{(n)} O\left(\frac{1}{n} \right) \leq \sum_{k=1}^{\sqrt{n}} O\left(\frac{1}{n} \right),$$ but the sum on the RHS can (theoretically) be unbounded, as $n\to \infty$.
I have to factor in the fact that I'm summing probabilities, and the sum is therefore bounded above by $1$, but how to do that?
If $$ \lim_{n \to \infty}\frac{a_k}{\frac{1}{n}} = c_1\\ \lim_{n \to \infty}\frac{r_k}{\frac{1}{n}} = c_2 $$ you get $$ \sum_{k=1}^{\sqrt{n}}a_k(1 \pm r_k) \leq\sum_{k=1}^{\sqrt{n}}\frac{c_1}{n}(1+ \frac{c_2}{n}) = \sum_{k=1}^{\sqrt{n}}\frac{c_1}{n} + \sum_{k=1}^{\sqrt{n}}\frac{c_1 c_2}{n^2} = \frac{c_1}{\sqrt{n}} + \frac{c_1 c_2}{n^{\frac{3}{2}}} = O(\frac{1}{\sqrt{n}}) $$ So your sum on the right is of the order $O(\frac{1}{\sqrt{n}})$