Compare-it-p-1-2-2-1-3-2-1-100-2-and-q-0-99-a-p-q-b-p-lt-q-c-p-gt-q-d-p-2-q-2-0-e-q-p-2-

Question Number 165829 by HongKing last updated on 09/Feb/22

Compare it :

p = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{100^{2}} and q = 0, 99

(a) p = q (b) p < q (c) p > q (d) p^{2} + q^{2} = 0

(e) \sqrt{q} = \sqrt{p} - 2

Answered by MJS_new last updated on 09/Feb/22

this is a weird question

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6} < \frac{{(\frac{22}{7})}^{2}}{6} = \frac{242}{147} \Rightarrow

\Rightarrow \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6} - 1 < \frac{95}{147} < \frac{98}{147} = \frac{2}{3} \Rightarrow

\Rightarrow \underset{n = 2}{\sum^{100}} \frac{1}{n^{2}} < \frac{2}{3} \Rightarrow

\Rightarrow p < q

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