If y n &gt; 0, lim x n y n = a and ∑ y n = + ∞, then lim x 1 + x 2 + . . . + x n y 1 + y 2 + . . . + y n = a

Question

If       y    n    &amp;gt;  0,   lim                    x        n                    y        n              =    a   and   ∑            y      n        =  +  ∞, then   lim                              x          1                +                  x          2                +        .        .        .        +                  x          n                                      y          1                +                  y          2                +        .        .        .        +                  y          n                      =    a

Trey Ross · Accepted Answer

You had a good idea. Indeed, since             x      n              y      n        →  a, then for any   ε  &amp;gt;  0 there is some N such that for   k  ≥  N you get   (  a  −  ε  )      y    k    ≤      x    k    ≤  (  a  +  ε  )      y    k   (we&#039;ve used that       y    k    &amp;gt;  0 here, so that the signs of inequalities remain). Then for   n  ≥  N you can rewrite                    x        1            +      .      .      .      +              x        n                            y        1            +      .      .      .      +              y        n              =                    x        1            +      .      .      .      +              x        N                            y        1            +      .      .      .      +              y        n              +                    x                  N          +          1                    +      .      .      .              x        N                            y        1            +      .      .      .      +              y        n              ≤                    x        1            +      .      .      .      +              x        N                            y        1            +      .      .      .              y        n              +  (  a  +  ε  )                    y                  N          +          1                    +      .      .      .      +              y        n                            y        1            +      .      .      .      +              y        n            Now, with N fixed, you have                     x        1            +      .      .      .      +              x        N                            y        1            +      .      .      .      +              y        n              →  0 as   n  →  ∞, due to divergence of series       ∑    k        y    k  . Moreover,                    y                  N          +          1                    +      .      .      .              y        n                            y        1            +      .      .      .      +              y        n              =  1  −                    y        1            +      .      .      .      +              y        N                            y        1            +      .      .      .      +              y        n                  →          n      →      ∞            1  −  0  =  1due to the same reason. Hence       lim sup    n                      x        1            +      .      .      .      +              x        n                            y        1            +      .      .      .      +              y        n              ≤  a  +  ε. Similarly you can prove that       lim inf    n                      x        1            +      .      .      .              x        n                            y        1            +      .      .      .      +              y        n              ≥  a  −  ε and since   ε  &amp;gt;  0 was arbitrary you can conclude  a  ≤      lim inf    n                      x        1            +      .      .      .      +              x        n                            y        1            +      .      .      .      +              y        n              ≤      lim sup    n                      x        1            +      .      .      .      +              x        n                            y        1            +      .      .      .      +              y        n              ≤  aand so the limit exists and is equal to a.

If y n </msub> > 0 , <mo movablelimits="true" form="prefix">lim <mrow clas

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