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It is asked to prove that the series $$\sum_n (\prod_{i=1}^n \frac{2i-1}{2i})^3$$

converges.

Unfortunately, the ratio test is not conclusive, so I am trying to apply the comparasion test. I've noted that

$$\prod (\frac{2i-1}{2i})^3 = (1/2^3)^n \prod (\frac{2i-1}{i})^3$$

Which not helps, since 2i -1 > i

I would be glad if someone could leave some suggestion.

Thanks in advance!

Giiovanna
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  • Can you prove that $\prod_{i=1}^{n}\frac{2i-1}{2i}=\theta\left(\frac{1}{\sqrt{n}}\right)$ by considering logarithms, for instance? – Jack D'Aurizio Sep 02 '14 at 20:47
  • I've tried to apply the log on the produtory but couldnt find it. – Giiovanna Sep 02 '14 at 20:52
  • $\log\left(1-\frac{1}{2i}\right) = -\frac{1}{2i}+O\left(\frac{1}{i^2}\right).$ By summing over $i$, then exponentiating back, you get that the product behaves like $\frac{1}{\sqrt{n}}$, hence the series is convergent. – Jack D'Aurizio Sep 02 '14 at 20:54

1 Answers1

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In my answer here I show that $$ \prod_{i=1}^n \frac{2i-1}{2i}\leq {3\over 4\sqrt{n+1}},$$ which gets you what you want.

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