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How to find a limit of: $$ lim_{n\to\infty}\frac{\sqrt[n]{\prod_{i=0}^{n-1} (a+ib)}}{\sum_{i=0}^{n-1} (a+ib)} $$ where $a>0$ and $b>0$

The task context is Infinite Product. Please help me or just give me a tip on how to simplify the expression and approach the task.

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  • $\begingroup$ I believe that the factor $\tfrac1n$ is missing in the denominator, and the expression should read $ \lim_{n\to\infty} {\frac{\sqrt[n]{\prod_{i=0}^{n-1}(a+ ib)}} {\tfrac1n\sum_{i=0}^{n-1}(a+ib)}} , $ that is, the ratio of geometric and arithmetic means of $n$ elements in arithmetic progression, which is indeed equal to $\tfrac{2}{e}$ as it was stated earlier in the recently deleted question. $\endgroup$
    – g.kov
    Sep 17, 2019 at 12:08
  • $\begingroup$ I have rechecked the task and there is no $\frac{1}{n}$ factor, maybe is a typo in the textbook. $\endgroup$ Sep 17, 2019 at 12:12
  • $\begingroup$ @g.kov But if You are right then this expression has a really practical use. $\endgroup$ Sep 17, 2019 at 12:14
  • $\begingroup$ The question would look more interesting, if you include your thoughts about this practical use in the body of the question. $\endgroup$
    – g.kov
    Sep 17, 2019 at 12:34

1 Answer 1

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Let use arithmetic and geometric mean inequality: $$ lim_{n\to\infty}\frac{\sqrt[n]{\prod_{i=0}^{n-1} (a+ib)}}{\sum_{i=0}^{n-1} (a+ib)} < lim_{n\to\infty}\frac{\sum_{i=0}^{n-1} (a+ib)/n}{\sum_{i=0}^{n-1} (a+ib)} = lim_{n\to\infty} \frac{1}{n} = 0 $$ Or equivalency we can replace the denominator: $$ lim_{n\to\infty}\frac{\sqrt[n]{\prod_{i=0}^{n-1} (a+ib)}}{\sum_{i=0}^{n-1} (a+ib)} < lim_{n\to\infty}\frac{\sqrt[n]{\prod_{i=0}^{n-1} (a+ib)}}{n\sqrt[n]{\prod_{i=0}^{n-1} (a+ib)}} = lim_{n\to\infty} \frac{1}{n} = 0 $$ Notice that the equality hold iff $n=1$.

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    $\begingroup$ Using logarithms, the expression is effectively $\frac{2}{e n}+O\left(\left(\frac{1}{n}\right)^2\right)$ $\endgroup$ Sep 17, 2019 at 11:49

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