Question-213642

Question Number 213642 by universe last updated on 12/Nov/24

Answered by Berbere last updated on 12/Nov/24

a_{n} = \lim_{N \to \infty} \underset{k = n}{\sum^{N}} \frac{1}{k^{2}}; \forall k ⩾ n > 1 k (k - 1) ⩽ k^{2} ⩽ k (k + 1) \Rightarrow

\lim_{N \to \infty} \underset{k = n}{\sum^{N}} \frac{1}{k (k - 1)} \underset{}{⩾} a_{n} ⩾ \underset{k = n}{\sum^{\infty}} \frac{1}{k (k + 1)} = \underset{k = n}{\sum^{\infty}} \frac{1}{k} - \frac{1}{k + 1} = \frac{1}{n}

\frac{1}{n - 1} ⩾ a_{n} ⩾ \frac{1}{n} \Rightarrow \lim_{n \to \infty} n . a_{n} = 1

n^{2} a_{n} ⩾ n \Rightarrow \lim_{n \to \infty} n^{2} a_{n} = + \infty

Commented by universe last updated on 12/Nov/24

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