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.
 A: Riemann Sum Approach
$$
\begin{align}
\lim_{n\to\infty}\sum_{r=1}^n\frac{6n}{9n^2-r^2}
&=\lim_{n\to\infty}\sum_{r=1}^n\frac{2}{1-\frac{r^2}{9n^2}}\frac1{3n}\tag1\\
&=\int_0^{1/3}\frac2{1-x^2}\,\mathrm{d}x\tag2\\
&=\int_0^{1/3}\left(\frac1{1-x}+\frac1{1+x}\right)\mathrm{d}x\tag3\\
&=\left.\log\left(\frac{1+x}{1-x}\right)\,\right|_0^{1/3}\tag4\\[6pt]
&=\log(2)\tag5
\end{align}
$$
Explanation:
$(1)$: divide numerator and denominator by $9n^2$
$(2)$: Riemann Sum with $x=\frac{r}{3n}$
$(3)$: partial fractions
$(4)$: integrate
$(5)$: evaluate
A: I assume you mean the limit as $n\to \infty$. You can write $\dfrac{1}{3n+r}$ as $\dfrac{1}{n}\cdot\dfrac{1}{3+\frac{r}{n}}$. Then the sum $ \displaystyle\sum_{r=1}^n \frac{1}{3n+r}$ becomes the Riemann sum approximating $ \displaystyle\int_0^1 \frac{1}{3+x} dx$, and this integral is $\ln 4- \ln 3$. The other sum is similar and the overall limit is indeed $\ln 2$.  
A: $S(n) = \underbrace{\sum\limits_{1 \le k \le n} \frac{1}{3n - k }}_{A} + \underbrace{\sum\limits_{1 \le k \le n} \frac{1}{3n + k}}_{B}$
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 = $H_{4n} - H_{3n} + H_{3n-1} - H_{2n-1} = H_{4n} - H_{3n-1} - \frac{1}{3n} + H_{3n-1} - H_{2n-1} = \boxed{H_{4n} - H_{2n-1} - \frac{1}{3n}}$
A: You can continue as follows
$$ S(n)= \sum_{r=1}^n \Bigl(\frac{1}{3n+r} + \frac{1}{3n-r}\Bigr)$$
$$ = \sum_{r=1}^n \Bigl(\frac{1}{3+\frac rn} + \frac{1}{3-\frac rn}\Bigr)\frac1n$$
Let $\frac rn=x \implies x=0 $ as $r\to 1$  and $x=1$ as $r\to n$, $\frac1n=dx$ as $n\to \infty$
$$  \displaystyle\lim_{n\to \infty} \sum_{r=1}^n \frac{6n}{9n^2-r^2}= \int_0^1\left(\frac{1}{3+x} + \frac{1}{3-x}\right)dx$$
$$=\left(\ln|3+x|-\ln|3-x|\right)_0^1$$
$$=\ln 4-\ln2=\ln 2$$
