When does ∑ 1 ( n ln ⁡ n ) a converge?For a &gt; 1, I used the limit comparison test as 1 ( n ln ⁡ n ) a 1 n a → 0 as n → ∞ so that ∑ 1 ( n ln ⁡ n ) a converges together with ∑ 1 n a and for a ≤ 1 2 notice that ∑ 1 ( n ln ⁡ n ) a &gt; ∑ 1 n 2 a and since RHS diverges we can see ∑ 1 ( n ln ⁡ n ) a diverges . But what happens if a ∈ ( 1 2 , 1 )? Or is there any other method to cover these cases on one step?

Question

When does   ∑      1          (      n      ln      ⁡      n              )        a             converge?For   a  &amp;gt;  1, I used the limit comparison test as                     1                  (          n          ln          ⁡          n                      )            a                                      1                  n          a                      →    0   as   n  →  ∞ so that       ∑          1              (        n        ln        ⁡        n                  )          a                     converges together with       ∑          1              n        a             and for   a  ≤      1    2   notice that       ∑          1              (        n        ln        ⁡        n                  )          a                      &amp;gt;    ∑          1              n                  2          a                     and since RHS diverges we can see       ∑          1              (        n        ln        ⁡        n                  )          a                     diverges . But what happens if   a  ∈  (      1    2    ,  1  )? Or is there any other method to cover these cases on one step?

Jason Petersen · Accepted Answer

Here is an approach. By the Cauchy condensation test we can study instead the series      ∑          n                  2              (        1        −        a        )        n                    n      a        =      ∑          n                  z      n              n      a        ,where   z  =      2          (      1      −      a      )      . The last series has the radius of convergence       |    z      |    &amp;lt;  1 which implies      2          1      −      a        &amp;lt;  1    ⟹    1  −  a  &amp;lt;  0    ⟹    a  &amp;gt;  1Cauchy Condensation Theorem:For a positive non-increasing sequence   f  (  n  ), the sum      ∑          n      =      1              ∞        f  (  n  )converges if and only if the sum      ∑          n      =      0              ∞            2          n        f  (      2          n        )converges.

imire37 · Accepted Answer

Lemma: A series   ∑      a    k   whose terms are positive and monotonously decreasing is convergent iff the series       ∑          k      =      1        ∞        2    k        a                  2        k             converges. Sketch of the proof :       a    i   is decreasing so       a    1    +      a    2    +  (      a    4    +      a    4    )  +  (      a    8    +      a    8    +      a    8    +      a    8    )  +  …  ≤      a    1    +      a    2    +  (      a    3    +      a    4    )  +  (      a    5    +      a    6    +      a    7    +      a    8    )  +  .  .Again,      a    1    +  (      a    2    +      a    3    )  +  (      a    4    +      a    5    +      a    6    +      a    7    )  +  ⋯  ≤      a    1    +  (      a    2    +      a    2    )  +  (      a    4    +      a    4    +      a    4    +      a    4    )  +  …Now applying this lemma, we need to find the condition for which  ∑            2      k                      (                  2          k                k        ln        ⁡        2        )            a      which is quite easy to find.

When does sum(1)/((n\ln n)^a) converge?

Open question

Answer & Explanation

New Questions in Algebra II