Prove ∑ n = 0 ∞ 1 x n a n − 1 &lt; ∞ for 0 ≤ a &lt; 1 and 1 &lt; x n + 1 ≤ x n ∀ n

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Prove       ∑          n      =      0        ∞        1                                        x            n                                a            n                              −      1        &amp;lt;  ∞ for   0  ≤  a  &amp;lt;  1 and   1  &amp;lt;      x          n      +      1        ≤      x          n        ∀  n

xxsailojaixxv5 · Accepted Answer

For any   a  &amp;lt;  1 there is an integer       n    0    (  a  ) such that       a          n        ≤      1    2   for each   n  ≥      n    o    (  a  ). Thus:      ∑          n      =      0              ∞            (                  a        n                              x          n                −                  a          n                      )       ≤         ∑          n      =      0                      n        0            (      a      )      −      1            (                  a        n                              x          n                −                  a          n                      )    +      ∑          n      =              n        0            (      a      )              ∞            (                  a        n                              x          n                −                  a          n                      )  Then using       a    n    ≤      1    2   for all n, gives             a      n                      x        n            −              a        n              ≤  2      a    n   for all   n  ≥      n    0    (  a  ), which in turn gives:      ∑          n      =      0                      n        0            (      a      )      −      1            (                  a        n                              x          n                −                  a          n                      )    +      ∑          n      =              n        0            (      a      )              ∞            (                  a        n                              x          n                −                  a          n                      )    ≤         ∑          n      =      0                      n        0            (      a      )      −      1            (                  a        n                              x          n                −                  a          n                      )    +      ∑          n      =              n        0            (      a      )              ∞        2      a    n    .As each       x    n   satisfies       x    n    &amp;gt;  1, it follows that       ∑          n      =      0                      n        0            (      a      )      −      1            (                  a        n                              x          n                −                  a          n                      )   is a finite number of terms each finite, and so this is finite for each   a  &amp;lt;  1. Also, as       ∑          n      =              n        0            (      a      )              ∞        2      a    n   is a geometric series, it follws that       ∑          n      =              n        0            (      a      )              ∞        2      a    n   is bounded for   a  &amp;lt;  1. Thus, it follows that       ∑          n      =      0                      n        0            (      a      )            (                  a        n                              x          n                −                  a          n                      )    +      ∑          n      =              n        0            (      a      )              ∞        2      a    n   is bounded, and thus, so is       ∑          n      =      0              ∞            (                  a        n                              x          n                −                  a          n                      )

Prove <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> n =

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New Questions in Calculus 2