u_n = Σ_(k=n+1) ^(2n) (1/k) and v_n = Σ_(k=n) ^(2n−1) (1/k) • show that u_n and v_n are adjacent use ln(x+1) ≤ x and x≤−ln(1−x) and • show that u_n ≤ Σ_(k=n+1) ^(2n) (ln(k)−ln(k−1)) hence deduce that u_n ≤ ln2 • show that v_n ≥ Σ_(k=n) ^(2n−1) (ln(k+1)−ln(k)) hence deduce that v


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Question Number 215979 by alcohol last updated on 23/Jan/25

$u_{n} = \underset{k = n + 1}{\sum^{2 n}} \frac{1}{k} a n d v_{n} = \underset{k = n}{\sum^{2 n - 1}} \frac{1}{k}$ $∙ s h o w t h a t u_{n} a n d v_{n} a r e a d j a c e n t$ $u s e l n (x + 1) ⩽ x a n d x ⩽ - l n (1 - x) a n d$ $∙ s h o w t h a t u_{n} ⩽ \underset{k = n + 1}{\sum^{2 n}} (l n (k) - l n (k - 1))$ $h e n c e d e d u c e t h a t u_{n} ⩽ l n 2$ $∙ s h o w t h a t v_{n} ⩾ \underset{k = n}{\sum^{2 n - 1}} (l n (k + 1) - l n (k))$ $h e n c e d e d u c e t h a t v_{n} ⩾ l n 2$
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