The base case is the claim that
[tex]\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}[/tex]
which reduces to
[tex]\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86[/tex]
which is true.
Assume that the inequality holds for n = k ; that
[tex]\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}[/tex]
We want to show if this is true, then the equality also holds for n = k + 1 ; that
[tex]\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}[/tex]
By the induction hypothesis,
[tex]\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}[/tex]
Now compare this to the upper bound we seek:
[tex]\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}[/tex]
because
[tex](2k+1)(k+2) > (2k+2)(k+1)[/tex]
in turn because
[tex]2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0[/tex]
