Rudin's PMA: Theorem 3.29 ProofIf p &gt; 1, ∑ n = 2 ∞ 1 n ( l o g n ) p converges; if p ≤ 1, the series diverges.Proof: The monotonicity of the logarithmic function implies that { l o g n } increases. Hence { 1 / n l o g n } decreases, and we can apply Theorem 3.27 to the series above; this leads us to the series ∑ k = 1 ∞ 2 k ⋅ 1 2 k ( l o g 2 k ) p = ∑ k = 1 ∞ 1 ( k l o g 2 ) p = 1 ( l o g 2 ) p ∑ k = 1 ∞ 1 k p and Theorem 3.29 follows from Theorem 3.28.I have two questions:(1) How can we get the decrease of { 1 / n l o g n } from the increase of { l o g n }? I didn't see any conncection between them.(2) The author said that we can apply Theorem 3.27 to the series. However, in order to apply Theorem 3.27, I think we need to show { 1 / n ( l o g n ) p } is decreasing. But I don't know how to do that.Theorem 3.27: Suppose a 1 ≥ a 2 ≥ ⋯ ≥ 0. Then the series ∑ n = 1 ∞ a n converges if and only if the series ∑ k = 0 ∞ 2 k a 2 k = a 1 + 2 a 2 + 4 a 4 + 8 a 8 + ⋯ converges.

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Rudin&#039;s PMA: Theorem 3.29 ProofIf   p  &amp;gt;  1,       ∑          n      =      2              ∞            1          n      (      l      o      g             n              )        p             converges; if   p  ≤  1, the series diverges.Proof: The monotonicity of the logarithmic function implies that   {  l  o  g     n  } increases. Hence   {  1      /    n     l  o  g     n  } decreases, and we can apply Theorem 3.27 to the series above; this leads us to the series       ∑          k      =      1              ∞            2    k    ⋅      1                  2        k            (      l      o      g                     2        k                    )        p              =      ∑          k      =      1              ∞            1          (      k      l      o      g             2              )        p              =      1          (      l      o      g             2              )        p                  ∑          k      =      1              ∞            1          k      p       and Theorem 3.29 follows from Theorem 3.28.I have two questions:(1) How can we get the decrease of   {  1      /    n     l  o  g     n  } from the increase of   {  l  o  g     n  }? I didn&#039;t see any conncection between them.(2) The author said that we can apply Theorem 3.27 to the series. However, in order to apply Theorem 3.27, I think we need to show   {  1      /    n     (  l  o  g     n      )    p    } is decreasing. But I don&#039;t know how to do that.Theorem 3.27: Suppose       a    1    ≥      a    2    ≥  ⋯  ≥  0. Then the series       ∑          n      =      1              ∞            a    n   converges if and only if the series       ∑          k      =      0              ∞            2    k        a                  2        k              =      a    1    +  2      a    2    +  4      a    4    +  8      a    8    +  ⋯ converges.

Jamir Young · Accepted Answer

Step 1The function   g  (  x  )  =  x      log    p    ⁡  (  x  ) increases and is positive in the interval   (  1  ,  ∞  ). From that, it follows that   f  (  x  )  =      1          x              log        p            ⁡      x       decreases on   (  1  ,  ∞  ).Step 2The convergence of the series can then be analyzed either by the integral test or Cauchy&#039;s condensation theorem, as you proposed in your problem.

opositor5t · Accepted Answer

Step 1- When   p  =  0 divergence is direct for in such case you get the harmonic series       ∑    n        1    n  .Step 2- For   p  &amp;lt;  0 notice that             log              −        p              n    ≥      log          −      p        ⁡  2      1    n   for all   n  ≥  2 and so, the series diverges in such cases.Step 3- For   p  &amp;gt;  0, as Jean Leider explained, your summand is increasing in which case you can apply Cauchy&#039;s condensation test as you suggested.

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