Let λ &gt; 0 be fixed, and a , b &gt; 0 positive real numbers. We have a series which is defined as ∑ n = 0 ∞ ( ∏ j = 1 n 1 1 + λ 1 a + b ( j − 1 ) ) .

Question

Let   λ  &amp;gt;  0 be fixed, and   a  ,  b  &amp;gt;  0 positive real numbers. We have a series which is defined as      ∑          n      =      0              ∞                  (            ∏          j      =      1              n            1          1      +      λ              1                  a          +          b          (          j          −          1          )                                )        .

Maximo Sweeney · Accepted Answer

it suffices to show that there exists a constant   α  &amp;gt;  1 such that             ∑              j        =        1            n        log    ⁡                  (              1    +    λ          1              a        +        b        (        j        −        1        )                            )              &amp;gt;    α    log    ⁡    n   for sufficiently large n But      log    ⁡                  (              1    +    λ          1              a        +        b        (        j        −        1        )                            )              ≈    λ          1              a        +        b        (        j        −        1        )             when j is large; and comparing             ∑              j        =        1            n        λ          1              a        +        b        (        j        −        1        )             to an integral shows that             ∑              j        =        1            n        λ          1              a        +        b        (        j        −        1        )              ∼          λ      b        log    ⁡    n  . So it suffices (and indeed is perhaps necessary as well) to take       b    &amp;lt;          1      λ

Let λ > 0 be fixed, and a , b > 0 positive real numbers.

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