I got stuck trying to solve this limit. lim n → ∞ 1 + 1 2 + 1 4 + . . . . + 1 2 n 1 + 1 3 + 1 9 + . . . + 1 3 n

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I got stuck trying to solve this limit.      lim          n      →      ∞                  1      +              1        2            +              1        4            +      .      .      .      .      +              1                  2                      n                                      1      +              1        3            +              1        9            +      .      .      .      +              1                  3                      n

behk0 · Accepted Answer

We see, that       (    1    +          1      2        +          1      4        +          1      8        +    ⋯    +          1              2                  n                      )   and       (    1    +          1      3        +          1      9        +    .    .    .    +          1              3                  n                      )   are the geometric series.Therefore,      lim          n      →      ∞            (    1    +          1      2        +          1      4        +          1      8        +    ⋯    +          1              2                  n                      )    =      ∑          n      =      0              ∞                  (              1        2            )        n    =      1          1      −              1        2              =      2    ,      lim          n      →      ∞            (    1    +          1      3        +          1      9        +    .    .    .    +          1              3                  n                      )    =      ∑          n      =      0              ∞                  (              1        3            )        n    =      1          1      −              1        3              =      3    2  We will have      lim          n      →      ∞                  (      1      +              1        2            +              1        4            +              1        8            +      ⋯      +              1                  2                      n                              )              (      1      +              1        3            +              1        9            +      .      .      .      +              1                  3                      n                              )        =      2          3      2        =      4    3    .

I got stuck trying to solve this limit. <munder> <mo movablelimits="true" form="prefix">lim

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