Compute ∑ k = 2 ∞ ( − 1 ) k ln ⁡ k k

Question

Compute      ∑          k      =      2              ∞                  (      −      1              )                  k                            ln        ⁡                  k                      k

lilao8x · Accepted Answer

Consider, for   s  &amp;gt;  1, the following auxiliary convergent series:      ∑          k      =      1        ∞    (  −  1      )    k              log      ⁡      (      k      )              k      s        =  −            d                      d            s            ∑          k      =      1        ∞    (  −  1      )    k        1          k      s        =  −            d                      d            s            (    (          2              1        −        s              −    1    )    ζ    (    s    )    )  The value of the series in question is obtained as a limit:      ∑          k      =      1        ∞    (  −  1      )    k              log      ⁡      (      k      )        k    =      lim          s      ↘      1            (          2              1        −        s              log    ⁡    (    2    )    ζ    (    s    )    +          ζ      ′        (    s    )    (    1    −          2              1        −        s              )    )  Since   ζ  (  s  )  =      1          s      −      1        +  γ  +      O    (  s  −  1  ), and       ζ    ′    (  s  )  =  −      1          (      s      −      1              )        2              +      O    (  1  ) we arrive at:      ∑          k      =      1        ∞    (  −  1      )    k              log      ⁡      (      k      )        k    =  γ  log  ⁡  (  2  )  −                    log        2            ⁡      (      2      )        2

Compute <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> k =

Answered question

Answer & Explanation

New Questions in Calculus 2