Question

\( \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} \) is equal to :

• A. \( {\log _e}\left( {{{\sqrt 3 } \over 2}} \right) \)
• B. \( {\log _e}\left( {{{3\sqrt 3 } \over 4}} \right) \)
• C. \( {\log _e}\left( {{{27} \over 4}} \right) \)
• D. \( {\log _e}\left( {{4 \over 3}} \right) \)

Hint:

Write the sum as \(\dfrac1n\sum f\left(\tfrac rn\right)\) and convert it to \(\int_0^1\dfrac{x}{2x^2-7x+6}dx\); use partial fractions.

Answer 1

The Correct Option is B

Step 1: Divide numerator and denominator by \(n^2\): the sum \(=\dfrac1n\sum_{r=1}^n\dfrac{r/n}{2(r/n)^2-7(r/n)+6}\). As \(n\to\infty\) this tends to \(\int_0^1\dfrac{x\,dx}{2x^2-7x+6}\).
Step 2: \(2x^2-7x+6=(2x-3)(x-2)\). Partial fractions: \(\dfrac x{(2x-3)(x-2)}=\dfrac{-3}{2x-3}+\dfrac2{x-2}\) (check: \(x=2\Rightarrow B=2\); \(x=\tfrac32\Rightarrow A=-3\)).
Step 3: Integral \(=\left[-\tfrac32\ln|2x-3|+2\ln|x-2|\right]_0^1\). At \(x=1\): \(0\). At \(x=0\): \(-\tfrac32\ln3+2\ln2\).
Step 4: Value \(=\tfrac32\ln3-2\ln2=\ln\dfrac{3^{3/2}}{4}=\ln\dfrac{3\sqrt3}4\).
Answer: (B) \(\log_e\left(\tfrac{3\sqrt3}4\right)\)