First slide

Special Series in Sequences and Series

Question

$If \sum T_{r} = \frac{n (n + 1) (n + 2)}{3} then lim_{n \to \infty} \sum_{r = 1}^{n} \frac{2008}{T_{r}} =$

Moderate

A

2008
B

8002
C

2080
D

2000

Solution

$\Rightarrow t_{n} = \sum_{r = 1}^{n} t_{r} - - \sum_{r = 1}^{n - 1} t_{r} = n (n + 1)$
$\begin{array}{l} = 2008 [\sum_{r = 1}^{n} (\frac{1}{r} - \frac{1}{r + 1})] \\ \cdot lim_{n \to \infty} \sum \frac{2008}{t_{n}} = lim_{n \to \infty} 2008 (1 - \frac{1}{n + 1}) = 2008 \end{array}$

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