# Asymptotics. Riemann sum?

I want to show that when suitable $k$ is chosen, this expression \begin{align*} \sum_{i=0}^{k} \frac{i}{2n - i} \end{align*} converges to some limit which does not involve $n$ as $n \to \infty$. I tried to use Riemann sum: because if $k = c n$ where $c$ is a constant \begin{align*} \sum_{i=0}^k \frac{1}{n} \frac{i/n}{2 - i/n} \to \int_0^c \frac{x}{2-x} \;\text{d} x. \end{align*} So $k = cn$ can stablize this summation when $n$ goes to infinity. But it is problematic: if $k = cn$ \begin{align*} \sum_{i=0}^{k} \frac{i}{2n - i} = n \sum_{i=0}^k \frac{1}{n} \frac{i/n}{2 - i/n} \to n \int_0^c \frac{x}{2-x} \;\text{d} x, \end{align*} which involves $n$. I would like to know if there is another method to achieve my goal.

Note: This is an intermediate step of a probability exercise. I showed that \begin{align*} \log \mathbb{P} (T_n > k) = - \sum_{i=0}^{k} \frac{i}{2n - i} + \text{small value (goes to 0 as $n \to \infty$)}. \end{align*} I want to find $a_n$ such that $T_n / a_n$ converges to some non-trivial random variable in distribution.

Edit: Thank you all! I think I have found the right order of $k$. @Jack and @Leonardo gave the same result through different approaches. I will update this post and mark the correct answer once I have the solution next week.

• Is it possible to tell us what is the original problem? – Jacky Chong Nov 12 '17 at 7:55
• To add to the answers below, if $k \gg \sqrt{n}$ then $\sum_{i=0}^{k} \frac{i}{2n-i} \to \infty$ as $n \to \infty$. – Antonio Vargas Nov 12 '17 at 16:35

$$\sum_{i=0}^{k}\frac{i}{2n-i}=2n\sum_{i=0}^{k}\frac{1}{2n-i}-\sum_{i=0}^{k}1=2n\left(H_{2n}-H_{2n-k-1}\right)-(k+1)$$ due to the asymptotic expansion $H_n=\log(n)+\gamma+\frac{1}{2n}+O\left(\frac{1}{n^2}\right)$, for large values of $n$ behaves like $\frac{k(k+1)}{4n}+O\left(\frac{k^3}{n^2}\right)$, hence any $k$ of the $\Theta(\sqrt{n})$ kind does the job.
• Thank you. But what is $H_n$? – jwyao Nov 12 '17 at 15:43
• @jwyao: the $n$-th harmonic number, $H_n=\sum_{h=1}^{n}\frac{1}{h}$. – Jack D'Aurizio Nov 12 '17 at 15:45
Try $k=k(n)=\sqrt{n}$. In this case, the limit of your sum is $\dfrac{1}{4}$:
$$\dfrac{k(n)(k(n)+1)}{2}\dfrac{1}{2n} = \sum \dfrac{i}{2n}\le \sum\dfrac{i}{2n-i}\le \sum \dfrac{i}{2n-\sqrt{n}} = \dfrac{k(n)(k(n)+1)}{2}\dfrac{1}{2n-\sqrt{n}}$$