Given that ( a n ) n = 1 ∞ is a sequence for which a n / log ⁡ ( n ) → 1 as n → ∞, show that lim x → 1 ∑ n = 0 ∞ a n x n 1 1 − x log ⁡ ( 1 1 − x ) = 1

Question

Given that   (      a    n        )          n      =      1              ∞       is a sequence for which       a    n        /    log  ⁡  (  n  )  →  1 as   n  →  ∞, show that      lim          x      →      1                          ∑                  n          =          0                          ∞                            a        n                    x                  n                                               1                  1          −          x                    log      ⁡              (                  1                      1            −            x                          )              =  1

Donavan Mack · Accepted Answer

1          1      −      x        log  ⁡      (          1              1        −        x              )    =      ∑          n      =      1        ∞        H    n        x    n    ,where       H    n    =  1  +      1    2    +  ⋯  +      1    n  . Note that   0  &amp;lt;      H    n    −  log  ⁡  n  ≤  1 holds. These are not so difficult to prove, so you should try doing so.Combined with       lim          n      →      ∞                  a      n              log      ⁡      n        =  1, we get       lim          n      →      ∞                  a      n              H      n        =  1 . So, for any   ε  &amp;gt;  0, there exists N so that   n  ≥  N implies       |              a      n              H      n        −  1      |    &amp;lt;  εFix N and let   f  (  x  )  =      ∑          n      =      0              N      −      1            a    n        x    n    −      ∑          n      =      1              N      −      1            H    n        x    n   for simplicity. Since N is a fixed finite number, note that   f  (  x  ) is bounded near   x  =  1                              |                                                    ∑                                  n                  =                  0                                                  ∞                                                            a                n                                            x                n                                                                    1                                  1                  −                  x                                            log              ⁡              (                              1                                  1                  −                  x                                            )                                −          1          |                                    =                  |                                                    ∑                                  n                  =                  0                                ∞                                            a                n                                            x                n                            −                              ∑                                  n                  =                  1                                ∞                                            H                n                                            x                n                                                                    ∑                                  n                  =                  1                                ∞                                            H                n                                            x                n                                              |                                                  ≤                                            |                        f            (            x            )                          |                                                          ∑                              n                =                1                            ∞                                      H              n                                      x              n                                      +                                            |                                      ∑                              n                =                N                            ∞                        (                          a              n                        −                          H              n                        )                          x              n                                      |                                                          ∑                              n                =                1                            ∞                                      H              n                                      x              n                                                        (by triangular inequality)                                          ≤                                            |                        f            (            x            )                          |                                                          ∑                              n                =                1                            ∞                                      H              n                                      x              n                                      +                                            ∑                              n                =                N                            ∞                                      |                                      a              n                        −                          H              n                                      |                                      x              n                                                          ∑                              n                =                1                            ∞                                      H              n                                      x              n                                                        (by triangular inequatlity)                                          ≤                                            |                        f            (            x            )                          |                                                          ∑                              n                =                1                            ∞                                      H              n                                      x              n                                      +        ε                                            ∑                              n                =                N                            ∞                                      H              n                                      x              n                                                          ∑                              n                =                1                            ∞                                      H              n                                      x              n                                                                  (by                                     |                                      a              n                        −                          H              n                                      |                        &amp;lt;            ε                          H              n                                )                                                  ≤                                            |                        f            (            x            )                          |                                                          ∑                              n                =                1                            ∞                                      H              n                                      x              n                                      +        ε            Recalling that f(x) is bounded near   x  =  1 implies the RHS   →  ε. Therefore we get      lim sup          x      →      1      −      0        LHS  ≤  εSince   ε is arbitrary positive number, we conclude that      lim          x      →      1      −      0        LHS  =  0.

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