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Question Number 136758 by mathlove last updated on 25/Mar/21

1 ==> \lim_{x \to \infty} (\frac{1}{π} + \frac{1}{π^{2}} + \frac{1}{π^{3}} + \cdot \cdot \cdot + \frac{1}{π^{n}}) = ?

2 => \lim_{x \to \infty} \frac{1 + 2^{2} + 3^{2} + \cdot \cdot \cdot \cdot + n^{2}}{1 - n^{3}} = ?

Answered by Dwaipayan Shikari last updated on 25/Mar/21

(1) \frac{1}{π} + \frac{1}{π^{2}} + \dots = \frac{\frac{1}{π}}{1 - \frac{1}{π}} = \frac{1}{π - 1}

(2) \lim_{n \to \infty} \frac{1 + 2^{2} + 3^{2} + . . + n^{2}}{1 - n^{3}} = \frac{n (n + 1) (2 n + 1)}{6 (1 - n^{3})} \sim - \frac{n (n + 1) (2 n + 1)}{6 n^{3}}

= - \frac{(1 + \frac{1}{n}) (2 + \frac{1}{n})}{6} = - \frac{1}{3}