JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

Let

P_n=\frac{2^3-1}{2^3+1} \cdot \frac{3^3-1}{3^3+1}+\cdots+\frac{n^3-1}{n^3+1} ; n=2,3,4, \ldots

Then

\lim \limits_{{n \rightarrow \infty}} P_n

is equal to

View answer and explanation

Correct option: D

\frac{2}{3}

Option D (

\frac{2}{3}

) is correct because after factoring and telescoping the product, the limit as n approaches infinity is

\frac{2}{3}

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