# Find a ratio of infinite product to infinite sum.

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.

• 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. Sep 17, 2019 at 12:08
• I have rechecked the task and there is no $\frac{1}{n}$ factor, maybe is a typo in the textbook. Sep 17, 2019 at 12:12
• @g.kov But if You are right then this expression has a really practical use. Sep 17, 2019 at 12:14
• The question would look more interesting, if you include your thoughts about this practical use in the body of the question. Sep 17, 2019 at 12:34

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