Does ∑ n = 1 ∞ ( ln ⁡ n ) 2 n 2 converge?

Question

Does       ∑          n      =      1              ∞                          (        ln        ⁡        n                  )          2                            n        2             converge?

apopihvj · Accepted Answer

There&#039;s one more way. Compare it to the integral:  I  =      ∫          1              ∞                  (                  log      ⁡      x        x                      )              2    d  xAlso note             log      ⁡      x              x      2        &amp;lt;            (                  log      ⁡      x        x                      )              2    &amp;lt;            log      ⁡      x        x    &amp;lt;                    log        2            ⁡      x        x  and the limit of all these sequences is 0 as   x  →  ∞. Now integrate I by parts in the following way:  I  =      ∫          0              ∞        1  ⋅            (                  log      ⁡      x        x                      )              2    d  x  =                    log        2            ⁡      x        x                      |                    1              ∞        −  2      ∫          1              ∞                  log      ⁡      x              x      2        d  x  +  2      ∫          1              ∞                  (                  log      ⁡      x        x                      )              2    d  x    =  −  2      ∫          1              ∞                  log      ⁡      x              x      2        d  x  +  2  Iso you get  I  =  2      ∫          1              ∞                  log      ⁡      x              x      2        d  xAgain integrate the expression on RHS by parts to get  I  =  2  (  0  +      ∫          1              ∞                  d      x              x      2        )  =  2Hence the original sum converges too.

inurbandojoa · Accepted Answer

To prove   ln  ⁡  n  &amp;lt;  C  (  ϵ  )      n          ϵ       where   C  (  ϵ  ) depends only on   ϵ:Since       e    x    ≥  1  +  x  &amp;gt;  x,   x  &amp;gt;  ln  ⁡  xSubstitute       x          ϵ       for x. Then       x          ϵ        &amp;gt;  ln  ⁡  (      x          ϵ        )  =      ϵ    ln  ⁡  x so   ln  ⁡  x  &amp;lt;                    x                  ϵ                            ϵ            Setting       ϵ    =      1    m  ,   ln  ⁡  x  &amp;lt;  m      x          1              /            m      . In particular, for   m  =  2,   ln  ⁡  x  &amp;lt;  2      x          1              /            2      .

Does sum_(n=1)^infty (ln n)^2/n^2 converge?

Answered question

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