Sequences & Series
Series sum using partial fractions; telescoping
MMTS_Full_Test_10
Grade 12
Question:
If $\displaystyle\sum_{r=1}^n T_r = \dfrac{n(n+1)(n+2)(n+3)}{12}$, where $T_r$ denotes the $r$-th term, then the value of $\displaystyle\lim_{n\to\infty}\sum_{r=1}^n \dfrac{1}{T_r}$ is
(A) $\dfrac{1}{4}$
(B) $\dfrac{1}{2}$
(C) $\dfrac{3}{4}$
(D) $1$
Step-by-Step Solution
Key Concept: Find $T_n$ by differencing: $T_n=S_n-S_{n-1}=\frac{n(n+1)(n+2)(n+3)}{12}-\frac{(n-1)n(n+1)(n+2)}{12}=\frac{n(n+1)(n+2)}{3}$. Then use partial fractions to sum $\sum 1/T_n$.
$T_n=\frac{n(n+1)(n+2)}{3}$. $\sum_{r=1}^\infty\frac{1}{T_r}=\frac{3}{2}\cdot\frac{1}{1\cdot2}=\frac{3}{4}$.
Correct Answer: (C) $\dfrac{3}{4}$