Given that ‘n’ is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction [tex] \displa

Question

Given that ‘n’ is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction
 \displaystyle \large{ \frac{1}{1}  +  \frac{1}{2}  +  \frac{1}{3}  + ... +  \frac{1}{n}  >  \frac{2n}{n + 1} }
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Josie 20 mins 2022-05-14T11:50:11+00:00 2 Answers 0 views 0

Answers ( )

    0
    2022-05-14T11:51:16+00:00

    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

    0
    2022-05-14T11:51:24+00:00

    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

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45:5+15-5:5+20+17*12-15 = ? ( )