How would you show that ∑ k = 1 ∞ ( log ⁡ k ) m k 1 + δ &lt; ∞for δ &gt; 0 and m ∈ N ?

Question

How would you show that      ∑          k      =      1        ∞              (      log      ⁡      k              )        m                    k              1        +        δ              &amp;lt;  ∞for   δ  &amp;gt;  0 and   m  ∈      N  ?

grcalia1 · Accepted Answer

Using L&#039;Hôpital&#039;s rule repeatedly, and being somewhat lazy:                              lim                      k            →            ∞                                                              k                              δ                                  /                                2                                                    (              log              ⁡              k                              )                m                                                                =                  lim                      k            →            ∞                                                                                                δ                  2                                                            k                                                                            δ                      2                                                        −                  1                                                                    m              (              log              ⁡              k                              )                                  m                  −                  1                                            ⋅                                                1                  k                                                                                              =                  C          1                          lim                      k            →            ∞                                                              k                                                δ                  2                                                                    (              log              ⁡              k                              )                                  m                  −                  1                                                                                              =                  C          2                          lim                      k            →            ∞                                                              k                                                δ                  2                                                                    (              log              ⁡              k                              )                                  m                  −                  2                                                                                                       ⋮                                                                     =                  C          m                          lim                      k            →            ∞                                                k                                          δ                2                                                    ,            for some positive constants       C    m  . Since       lim          k      →      ∞                  k                        δ          2                      =  ∞, it follows that       lim          k      →      ∞                          k                  δ                      /                    2                            (        log        ⁡        k                  )          m                      =  ∞  . Consequently, there is an N so that for   k  ≥  N, we have   (  log  ⁡  k      )    m    ≤      k                  δ        2            Then, for any   n  ≥  N and nonnegative integer m:  0  &amp;lt;      ∑          k      =      n              n      +      m                          (        log        ⁡        k                  )          m                            k                  1          +          δ                      ≤      ∑          k      =      n              n      +      m                          k                              δ            2                                      k                  1          +          δ                      =      ∑          k      =      n              n      +      m                  1              k                  1          +          δ                      /                    2                                      ⟶              n        ,        m        →        ∞                    0  ;whence the result follows.

How would you show that <munderover> ∑ <mrow class="MJX-TeXAtom-ORD">

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