Question

Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction

Answer 1

The base case is the claim that

$\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}$

which reduces to

$\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86$

which is true.

Assume that the inequality holds for n = k ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}$

We want to show if this is true, then the equality also holds for n = k + 1 ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}$

By the induction hypothesis,

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}$

Now compare this to the upper bound we seek:

$\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}$

because

$(2k+1)(k+2) > (2k+2)(k+1)$

in turn because

$2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0$

Answer 2

Answer:

1

11

11

11 +

11 + 2

11 + 21

11 + 21

11 + 21 +

11 + 21 + 3

11 + 21 + 31

11 + 21 + 31

11 + 21 + 31 +...+

11 + 21 + 31 +...+ n

11 + 21 + 31 +...+ n1

11 + 21 + 31 +...+ n1

11 + 21 + 31 +...+ n1 >

11 + 21 + 31 +...+ n1 > n+1

11 + 21 + 31 +...+ n1 > n+12n

11 + 21 + 31 +...+ n1 > n+12n

11 + 21 + 31 +...+ n1 > n+12n

Step-by-step explanation:

espero ter ajudado e bons estudos