How to solve the inequality n ! ≤ n n − 2 ?

Question

How to solve the inequality   n  !  ≤      n          n      −      2      ?

nuvolor8 · Accepted Answer

Here&#039;s an elementary proof, with some details left out to be filled in by you :).Taking logarithms, you want to show that       S    n    =  log  ⁡  2  +  log  ⁡  3  +  …  log  ⁡  n  ≤  (  n  −  2  )  log  ⁡  n    .Note that there are   n  −  1 terms on the left, each less than   log  ⁡  n. So you only have to &quot;squeeze out&quot; an additional term   log  ⁡  n. This can be done as follows:Let   k  =  ⌊      n    ⌋ , that is, k is the largest integer such that   k  ≤      n    &amp;lt;  k  +  1. Split the sum on the left to obtain:  log  ⁡  2  +  log  ⁡  3  +  ⋯  +  log  ⁡  k  +  log  ⁡  (  k  +  1  )  +  ⋯  +  log  ⁡  nThe terms up to   log  ⁡  k are all   ≤  log  ⁡      n    =      1    2    log  ⁡  n and there are   k  −  1 such terms. The remaining terms are all   ≤  log  ⁡  k  and there are   n  −  k such terms.Add everything up. The result is       S    n    ≤  (  k  −  1  )            log      ⁡      n        2    +  (  n  −  k  )  log  ⁡  n  =      (    n    −                  k        +        1            2        )    log  ⁡  n  .Now determine for which k and therefore for which n this implies       S    n    ≤  (  n  −  2  )  log  ⁡  n. You&#039;ll notice that you have proved the inequality for   n  ≥  9 or so. As Robert Israel already pointed out, the inequality actually holds for   n  ≥  5. These remaining cases can be checked by direct computation.

Leland Morrow · Accepted Answer

Here&#039;s a proof by induction.First, we need to verify the base case:   5  !  =  120  &amp;lt;  125  =      5    3  Now assume the induction hypothesis   n  !  &amp;lt;      n          n      −      2       is true, then we need to prove  (  n  +  1  )  !  &amp;lt;  (  n  +  1      )          n      −      1        ⇔  n  !  &amp;lt;  (  n  +  1      )          n      −      2        .But by induction hypothesis and the fact   n  &amp;lt;  n  +  1 we have,  n  !  &amp;lt;      n          n      −      2        &amp;lt;  (  n  +  1      )          n      −      2        .Therefore   (  n  +  1  )  !  &amp;lt;  (  n  +  1      )          n      −      1       and we are done.

How to solve the inequality n ! ≤ n <mrow class="MJX-TeXAtom-ORD"

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