$\lim_{n\rightarrow \infty}\prod^{n}_{r=1}(\sqrt{2}-2^{\frac{1}{(2r+1)}})$

Question

$$ \lim_{n\rightarrow \infty}\prod^{n}_{r=1}(\sqrt{2}-2^{\frac{1}{(2r+1)}})$$

When $n\rightarrow \infty $ Then last term i. e $(\sqrt{2}-2^{\frac{1}{(2n+1)}})\rightarrow (\sqrt{2}-1)$

Could some help me to solve it

Answer 1

We have that

\begin{align} \lim_{n\rightarrow \infty}\prod^{n}_{r=1}(\sqrt{2}-2^{\frac{1}{(2r+1)}}) &=\exp\left(\lim_{n\to\infty}\,\sum_{r=1}^n\,\log\left(\sqrt{2}-2^{\frac{1}{(2r+1)}}\right)\right)\\ &=\exp\left(\sum_{r=1}^{\infty}\,\log\left(\sqrt{2}-2^{\frac{1}{(2r+1)}}\right)\right), \end{align}

provided the limit exists of course. For a series $\sum a_n$ to converge, we must have $\lim_{n\to\infty} a_n=0$. This clearly fails here: as $r\to\infty$ we have

$$\log\left(\sqrt{2}-2^{\frac{1}{(2r+1)}}\right)\to\log\left(\sqrt{2}-1\right)<0,$$

so the series diverges to $-\infty$.

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In general, calculating an infinite product $\prod_{n=1}^\infty c_n$ is closely related to the series $\sum_{n=1}^\infty \log c_n$. The argument shows that in order for some infinite product to converge to a nonzero limit we must have $\lim_{n\to\infty}c_n=1$.

Much like the case for series, this is a necessary but not sufficient condition. If $\log c_n \to 0$, we will have $c_n \to 1$, but $\sum_{n=1}^\infty \log c_n$ may still fail to converge. In this case, $\prod_{n=1}^\infty c_n$ will also fail to converge.

Answer 2

You've got it. $0 < \sqrt 2 -1 < 1,$ so the limit will be $0$. For example, after some point, the factors will all be $< 1/2,$ so it doesn't matter what the factors before that point were.

I'll just point out that by the usual definition of infinite products, we would not say $$\prod_{r=1}^{\infty}{\sqrt2 - 2^{\frac{1}{2r+1}}} = 0.$$ The definition is adjusted to make it true that an infinite product converges $0$ if and only if one of the factors is $0$. The technical details are unimportant, and it is certainly true that the limit of the partial products, as shown in your question, converges to $0$.

Answer 3

Note that

$$0\le \prod^{n}_{r=1}(\sqrt{2}-2^{\frac{1}{(2r+1)}})\le\prod^{n}_{r=1}(\sqrt{2}-1)=(\sqrt{2}-1)^n\to0$$

thus by squeeze theorem the limit is 0.