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Question Number 218636 by hardmath last updated on 13/Apr/25

	Answered by vnm last updated on 14/Apr/25

	$l e t$ $a_{k} = \frac{1}{3 k - 1} + \frac{1}{3 k - 2} - \frac{2}{3 k}$ $b_{k} = \frac{4 k}{4 k + 1} - \frac{4 (k - 1)}{4 (k - 1) + 1}$ $i f \forall k ⩾ 1 a_{k} > b_{k} t h e n$ $\forall n ⩾ 1 \underset{k = 1}{\sum^{n}} a_{k} > \underset{k = 1}{\sum^{n}} b_{k} = \frac{4 n}{4 n + 1}$ $a_{k} = \frac{9 k - 4}{3 (3 k - 2) (3 k - 1) k}$ $b_{k} = \frac{4}{(4 k + 1) (4 k - 3)}$ $a_{k} - b_{k} = \frac{9 k - 4}{3 (3 k - 2) (3 k - 1) k} - \frac{4}{(4 k + 1) (4 k - 3)} =$ $\frac{36 k^{3} - 28 k^{2} - 19 k + 12}{3 k (3 k - 2) (3 k - 1) (4 k + 1) (4 k - 3)}$ $36 k^{3} - 28 k^{2} - 19 k + 12 = [2 k = m] =$ $\frac{1}{2} (9 m^{3} - 14 m^{2} - 19 m + 24) =$ $\frac{1}{2} (m - 1) (9 m^{2} - 5 m - 24) = 0$ $m_{1} = 1, m_{2, 3} = \frac{5 \pm \sqrt{889}}{18}$ $\frac{5 + \sqrt{889}}{18} < \frac{5 + 30}{18} < 2 \Rightarrow \forall k ⩾ 1 a_{k} - b_{k} > 0$
	Commented by hardmath last updated on 17/Apr/25

	$thankyou dearprofessor$
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