For instance, let the function: f = ∏ n = 1 ∞ a b ,for b ∈ N , b &gt; 1 and a irrational such that f converges to the real number k. We immediately see that all partial products are irrational. Can we then also say that k is irrational?

Question

For instance, let the function:  f  =      ∏          n      =      1              ∞            a    b    ,for   b  ∈      N  ,   b  &amp;gt;  1 and   a irrational such that   f converges to the real number   k. We immediately see that all partial products are irrational. Can we then also say that   k is irrational?

Kiana Cantu · Accepted Answer

Let   a  =      2   for all   n and       b    n    =  2 for   n  =  1  ,  2  ,  3 and       b    n    =      2          n      −      2       for   n  ≥  3Then you have       b    n   as the sequence   2  ,  2  ,  2  ,  4  ,  8  ,  16  ,  … and   ∑      1          b      n        =  2So      ∏          n      =      1              ∞            a          b      n                            =      a          ∑              1                  b          n                                                      =            2        2    =  2which is an integer limit despite all the       a          b      n                           being irrational

ttyme411gl · Accepted Answer

Let   p be a prime number and   n  ∈      N  . Clearly,       p          1              /            n       is irrational.Now, consider a sequence   (      n    i        )          i      ∈              N             such that       ∑          i      =      1        ∞        1          n      i        =  1.Observe that       ∏          i      =      1        ∞        p          1              /                    n        i              =  p.

For instance, let the function: f = <munderover> ∏ <mrow class

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New Questions in Algebra I