Find A = ∑ n = 1 ∞ log ⁡ 2 n 3 n

Question

Find   A  =      ∑          n      =      1              ∞                  log      ⁡              2        n                    3      n

Envetenib8ne · Accepted Answer

If the general term is   log  ⁡  (      2    n    )      /        3    n  , you have      ∑          n      =      1        ∞              log      ⁡      (              2        n            )              3      n        =  log  ⁡  (  2  )      ∑          n      =      1        ∞        n          3      n      Because 1/3&amp;lt;1, this series can be evaluated using the following formula:      ∑          n      =      1        ∞    n      z    n    =      z          (      1      −      z              )        2            This formula is derived by differentiating the geometric series (and multiplying by z), which brings us to:If the general term is   (  log  ⁡  (  2  )      )    n        /        3    n  , what you have is exactly a geometric series:      ∑          n      =      1        ∞              (      log      ⁡      (      2      )              )        n                    3      n        =      ∑          n      =      1        ∞              (                        log          ⁡          (          2          )                3            )        n  since   log  ⁡  (  2  )  &amp;lt;  3, we know that   log  ⁡  (  2  )      /    3  &amp;lt;  1, and so this series can be evaulated using the formula      ∑          n      =      1        ∞        z    k    =      z          1      −      z

Find A=sum_(n=1)^infty log 2^n/3^n

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