Question

Challenge :

For any natural number n, prove the following inequality :

\(\large \mathsf{ \cfrac{1}{n+1 } (1+\cfrac{1}{3} + \cfrac{1}{5} + ...+ \cfrac{1}{2n-1} ) \ge \cfrac{1}{n} (\cfrac{1}{2} + \cfrac{1}{4} + ... +\cfrac{1}{2n} ) }\)

Answer 1

1 < 2^1, so the proposition is true for n = 1.

Assume it's true for some natural number k > 0.

Then k < 2^k. Also 1 < 2^1 < 2^k.

Adding k < 2^k and 1 < 2^k gives

k + 1 < 2^k + 2^k = 2* (2^k) = 2^(k+1).

Answer 2

Yes , but how does that disprove the given inequality Jack ?

Answer 3

@FutureMathProfessor , how does that prove the given inequality.

Answer 4

We are talking about natural numbers here Jack.

\(\cfrac{1}{x} \) can not be considered here unless (unless x = 1 as it becomes a natural number then)

Answer 5

ah, my bad
i assumed that natural numbers meant whole intergers