<s>[INST] Finding $\lim_{n\to \infty} \sum_{r=1}^n \frac{6n}{9n^2-r^2}$ This is a question from an entrance exam for a college here in India.
We have to find $$ \lim_{n\to\infty}\sum_{r=1}^n \frac{6n}{9n^2-r^2}$$
So far I have tried:
$$ S(n)= \sum_{r=1}^n \frac{6n}{9n^2-r^2}$$
$$ S(n)= \sum_{r=1}^n \Bigl(\frac{1}{3n+r} + \frac{1}{3n-r}\Bigr)$$
from what I can notice as $n$ gets larger $\frac{1}{3n}$ gets closer to $0$ and so does the neighborhood of summation if I may call it, $\frac{1}{2n}$ and $\frac{1}{4n}$. So shouldn't it be closer to $0$. The answer key suggests that its $\log{2}$. I'm confused after the second step. Thanks for the answers.
 [/INST] $S(n) = \underbrace{\sum\limits_{1 \le k \le n} \frac{1}{3n - k }}_{A} + \underbrace{\sum\limits_{1 \le k \le n} \frac{1}{3n + k}}_{B}$
Let's denote $H_{n} = \sum\limits_{1 \le k \le n} \frac{1}{k}$ - harmonic series (we assume, that it's a known series).
Than A = substitution $\{ k = 3n - k  \} \rightarrow \sum\limits_{1 \le 3n-k \le n} \frac{1}{3n - (3n - k)} = \sum\limits_{2n \le k \le 3n - 1} \frac{1}{k} = \sum\limits_{1 \le k \le 3n-1}\frac{1}{k} - \sum\limits_{1 \le k \le 2n-1} \frac{1}{k} = H_{3n-1} - H_{2n-1}$.
B = substitution $\{ k = -3n + k  \} \rightarrow \sum\limits_{1 \le -3n + k \le n} \frac{1}{k} = \sum\limits_{3n+1 \le k \le 4n} \frac{1}{k} = \sum\limits_{1 \le k \le 4n} \frac{1}{k} - \sum\limits_{1 \le k \le 3n} \frac{1}{k} = H_{4n} - H_{3n}$
So: A + B = $H_{4n} - H_{3n} + H_{3n-1} - H_{2n-1} = H_{4n} - H_{3n-1} - \frac{1}{3n} + H_{3n-1} - H_{2n-1} = \boxed{H_{4n} - H_{2n-1} - \frac{1}{3n}}$
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