Determine the convergence or divergence of ∑ 2 ∞ 1 ( log ⁡ n ) s , where s ∈ R is given.Since 1 ( log ⁡ n ) s &gt; 1 n s for large n, if s ≤ 1 then ∑ 2 ∞ 1 ( log ⁡ n ) s diverges.But for s &gt; 1 I have not yet figured out a proof.

Question

Determine the convergence or divergence of       ∑          2              ∞            1          (      log      ⁡      n              )                  s                    , where   s  ∈      R   is given.Since      1          (      log      ⁡      n              )                  s                      &amp;gt;      1          n              s            for large   n, if   s  ≤  1 then       ∑          2              ∞            1          (      log      ⁡      n              )                  s                     diverges.But for   s  &amp;gt;  1 I have not yet figured out a proof.

Lena Gomez · Accepted Answer

A different proof.      lim          n      →      ∞                  (      log      ⁡      n              )        s              n    =  0    ∀  s  &amp;gt;  0.Thus      1          (      log      ⁡      n              )        s              ≥      1    n          for           n         large enough (depending on           s        .)

Brooke Richard · Accepted Answer

There is a theorem that states that if       a    n   is a decreasing sequence, then       ∑    n        a    n   converges iff       ∑    i        2    i        a                  2        i             converges.See chapter 3 of Baby Rudin for the proof.Applying that here we get      ∑    i              2      i              (      log      ⁡      (              2        i            )              )        s              =      1          (      log      ⁡      2              )        s                  ∑    i              2      i              i      s      which diverges, hence the original series diverges.

Determine the convergence or divergence of sum_(2)^oo 1/((log n)^s), where s in RR

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