prove-by-reccurence-1-2-1-4-1-6-1-2n-lt-or-n-2-thanks-

Question Number 108371 by Skabetix last updated on 16/Aug/20

p r o v e b y r e c c u r e n c e

\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots + \frac{1}{2 n} < o r = \frac{n}{2}

t h a n k s

Answered by Aziztisffola last updated on 16/Aug/20

for n = 1 \Rightarrow \frac{1}{2} ⩽ \frac{1}{2} true

suppose \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots + \frac{1}{2 n} ⩽ \frac{n}{2}

and prove that \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots + \frac{1}{2 n} + \frac{1}{2 n + 2} ⩽ \frac{n + 1}{2}

\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots + \frac{1}{2 n} + \frac{1}{2 n + 2} ⩽ \frac{n}{2} + \frac{1}{2 n + 2}

\frac{n}{2} + \frac{1}{2 n + 2} = \frac{n}{2} + \frac{1}{2 (n + 1)} ⩽ \frac{n}{2} + \frac{1}{2} = \frac{n + 1}{2}

\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots + \frac{1}{2 n} + \frac{1}{2 n + 2} ⩽ \frac{n + 1}{2}

hence \forall n \in N^{*} \underset{k = 1}{\sum^{n}} \frac{1}{2 k} ⩽ \frac{n}{2}

Answered by 1549442205PVT last updated on 16/Aug/20

i) For n = 1 we have \frac{1}{2} = \frac{1}{2} \Rightarrow true

ii) Consider n ⩾ 5 we have

\frac{1}{10} < \frac{1}{4.5} = \frac{1}{4} - \frac{1}{5}, \frac{1}{12} < \frac{1}{5.6} = \frac{1}{5} - \frac{1}{6} \dots

\frac{1}{2 n} < \frac{1}{(n - 1) n} = \frac{1}{n - 1} - \frac{1}{n}

\Rightarrow \frac{1}{10} + \frac{1}{12} + \dots + \frac{1}{2 n} < \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \dots

\dots + \frac{1}{n - 1} - \frac{1}{n} = \frac{1}{4} - \frac{1}{n} . Hence,

Adding up we get

LHS < (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8}) + \frac{1}{4} - \frac{1}{n} < \frac{n}{2}

\Leftrightarrow \frac{12 + 6 + 4 + 3}{24} - \frac{1}{n} < \frac{n}{2} \Leftrightarrow \frac{25}{24} - \frac{1}{n} < \frac{n}{2}

\Leftrightarrow \frac{25 n - 24}{24 n} < \frac{n}{2} \Leftrightarrow {12 n}^{2} - 25 n + 24 > 0

\Leftrightarrow 12 {(n - \frac{25}{24})}^{2} + 24 - \frac{625}{48} > 0

is true \forall n ⩾ 2

Consequently,

\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots + \frac{1}{2 n} ⩽ \frac{n}{2} (q . e . d)