Prove-that-n-N-H-2-n-1-n-2-where-H-m-r-1-m-1-r-

Question Number 2112 by Yozzi last updated on 03/Nov/15

P r o v e t h a t, \forall n \in N,

H (2^{n}) ⩾ 1 + \frac{n}{2}

w h e r e H (m) = \underset{r = 1}{\sum^{m}} \frac{1}{r} .

Commented by 123456 last updated on 03/Nov/15

n = 0 \Rightarrow H (2^{n}) = H (1) = 1 ⩾ 1 + \frac{0}{2}

n = 1 \Rightarrow H (2^{n}) = H (2) = \frac{3}{2} ⩾ 1 + \frac{1}{2}

Answered by 123456 last updated on 03/Nov/15

H (m) = 1 + \frac{1}{2} + \cdot \cdot \cdot + \frac{1}{m}

< 1 + \frac{1}{2} + \cdot \cdot \cdot + \frac{1}{2} m > 2 \Rightarrow \frac{1}{m} < \frac{1}{2}

< 1 + \frac{m - 1}{2}

H (m) > 1 + \frac{1}{m} + \cdot \cdot \cdot + \frac{1}{m}

> 1 + \frac{m - 1}{m}

1 + \frac{m - 1}{m} < H (m) < 1 + \frac{m - 1}{2}

- - - - - - -

continue

Answered by prakash jain last updated on 04/Nov/15

H (1) = 1

H (2^{1}) = 1 + \frac{1}{2} = \frac{3}{2} ⩾ \frac{1 + 1}{2}

H (2^{2}) = H (2) + \frac{1}{3} + \frac{1}{4} ⩾ \frac{3}{2} + \frac{1}{4} + \frac{1}{4} = \frac{3}{2} + \frac{1}{2} = \frac{1 + 3}{2}

assume H (2^{n}) ⩾ \frac{1 + n}{2}

H (2^{n}) = H (2^{n}) + \frac{1}{2^{n} + 1} + \frac{1}{2^{n} + 2} + \dots + \frac{1}{2^{n + 1}}

⩾ \frac{1 + n}{2} + \frac{1}{2^{n + 1}} + . . + \frac{1}{2^{n + 1}} = \frac{1 + n}{2} + \frac{2^{n}}{2^{n + 1}}

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

H (2^{n}) ⩾ \frac{1 + n}{2} \Rightarrow H (2^{n + 1}) ⩾ \frac{1 + (n + 1)}{2}

Result follows by induction .