Suppose: ∑ n = 2 ∞ ( 1 n ( ln ⁡ ( n ) ) k ) = 1 2 ( ln ⁡ ( 2 ) ) k + 1 3 ( ln ⁡ ( 3 ) ) k + . . . ,

Question

Suppose:      ∑          n      =      2              ∞            (          1              n        (        ln        ⁡        (        n        )                  )                      k                                )    =      1          2      (      ln      ⁡      (      2      )              )                  k                      +      1          3      (      ln      ⁡      (      3      )              )                  k                      +  .  .  .  ,

Erick Blake · Accepted Answer

For completeness, we sketch the Integral Test approach.Let f be a function which is defined, non-negative, and decreasing (or at least non-increasing from some point a on. Then       ∑    1    ∞    f  (  n  ) converges if and only if the integral       ∫    a    ∞    f  (  x  )    d  x converges.In our example, we use   f  (  x  )  =            1              x                          ln          k                ⁡        (        x        )            . Note that   f  (  x  )  ≥  0 after a while, and decreasing So we want to find the values of k for which      ∫    2    ∞              d      x              x              ln        k            ⁡      (      x      )            (  ∗  )converges (we could replace 2 by say 47 if f(x) misbehaved early on).Let   I  (  M  )  =      ∫    0    M    f  (  x  )    d  x. We want to find the values of k for which       lim          M      →      ∞        I  (  M  ) exists.Suppose first that   k  &amp;gt;  1. To evaluate the integral, make the substitution   ln  ⁡  x  =  u. Then  I  (  M  )  =      ∫    2    M              d      x              x              ln        k            ⁡      (      x      )        =      ∫          ln      ⁡      2              ln      ⁡      M                  d      u              u      k        .We find that  I  (  M  )  =      1          k      −      1            (          1              (        ln        ⁡        2                  )                      k            −            1                                −          1              (        ln        ⁡        M                  )                      k            −            1                                )    .Because   k  −  1  &amp;gt;  0, the term in M approaches 0 as   M  →  ∞, so the integral (∗) converges.By the Integral Test, we therefore conclude that our original series converges if   k  &amp;gt;  1For   k  =  1, after the u-substitution, we end up wanting   ∫            d      u        u  . We find that  I  (  M  )  =  ln  ⁡  (  ln  ⁡  M  )  −  ln  ⁡  (  ln  ⁡  2  )  .As   M  →  ∞ (though glacially slowly). So by the Integral Test, our series diverges if   k  =  1For   k  &amp;lt;  1, we could again do the integration. But an easy Comparison with the case   k  =  1 proves divergence.

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