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

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⇒tn=∑r=1ntr−−∑r=1n−1tr=nn+1=2008∑r=1n 1r−1r+1⋅limn→∞ ∑2008tn=limn→∞ 20081−1n+1=2008

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If ∑Tr=n(n+1)(n+2)3 then limn→∞ ∑r=1n 2008Tr=

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