Upper bound for a series r ∈ ( 0 , 1 ) for r ∈ ( 0 , 1 )

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Upper bound for a series   r  ∈  (  0  ,  1  ) for   r  ∈  (  0  ,  1  )

scipionhi · Accepted Answer

A crude bound is  f  (  x  )  &amp;lt;      x          (      1      −      x              )        2               for every   x  ∈  (  0  ,  1  )This can be obtained by noticing that our series is       ∑          n      =      0        ∞    n      x    n   minus the terms whose index is not a power of two. For   x  ∈  (  0  ,  1  ), the terms of       ∑          n      =      0        ∞    n      x    n   are all strictly positive and the series converges, so since our series contains less terms than       ∑          n      =      0        ∞    n      x    n  , the following inequality holds:      ∑          n      =      0        ∞        2    n        x          (              2        n            )        &amp;lt;      ∑          n      =      0        ∞    n      x    n  The sum of       ∑          n      =      0        ∞    n      x    n   is       x          (      1      −      x              )        2            , a fact that can be proven by differentiating the geometric series identity       ∑          n      =      0        ∞    =      1          1      −      x      , giving us the desired bound.  f  (  x  )  &amp;lt;      x          (      1      −      x              )        2

Upper bound for a series r ∈<!--

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