# how you solve: lim_(n->oo)(2-sqrt2)(2-root(3)2)(2-root(4)2)...(2-root(n)2)) ?

Mar 18, 2017

${\prod}_{k = 2}^{\infty} \left(2 - \sqrt[k]{2}\right) = 0$

#### Explanation:

At first glance, we would propose $0$ as the limit value. This conclusion is correct but must be proved.

So calling ${a}_{n} = 2 - \sqrt[n]{2}$, the condition ${a}_{n} < 1$ is not sufficient for convergence to zero in

${\lim}_{n \to \infty} {\prod}_{k = 2}^{n} \left(2 - {2}^{\frac{1}{k}}\right)$

The necessary and sufficient condition for convergence of

${\prod}_{k = 2}^{\infty} {a}_{k}$ to a non zero value is that ${\sum}_{k = 2}^{\infty} \log \left({a}_{k}\right)$ converges.

If ${a}_{n} < 1$ and ${\sum}_{k = 2}^{\infty} \log \left({a}_{k}\right)$ does not converge, then
${\prod}_{k = 2}^{\infty} {a}_{n}$ converges to $0$.

Now considering the inequality $\frac{x - 1}{x} \le \log \left(x\right) , x > 1$ we have

$\frac{1 - {2}^{\frac{1}{k}}}{2 - {2}^{\frac{1}{k}}} \le \log \left(2 - {2}^{\frac{1}{k}}\right)$

Considering now

${\sum}_{k = 2}^{\infty} \frac{1 - {2}^{\frac{1}{k}}}{2 - {2}^{\frac{1}{k}}}$

with ${b}_{k} = \frac{1 - {2}^{\frac{1}{k}}}{2 - {2}^{\frac{1}{k}}}$ clearly this series does not converge because ${b}_{k} > 0$ and

${\lim}_{k \to \infty} {b}_{k} / {b}_{k + 1} = 1$

so because ${\sum}_{k = 2}^{\infty} \frac{1 - {2}^{\frac{1}{k}}}{2 - {2}^{\frac{1}{k}}} \le {\sum}_{k = 2}^{\infty} \log \left(2 - {2}^{\frac{1}{k}}\right)$

${\sum}_{k = 2}^{\infty} \log \left(2 - {2}^{\frac{1}{k}}\right)$ does not converge then finally

${\lim}_{n \to \infty} {\prod}_{k = 2}^{n} \left(2 - {2}^{\frac{1}{k}}\right) = 0$

Mar 18, 2017

${\prod}_{n = 2}^{\infty} \left(2 - {2}^{\frac{1}{n}}\right) = 0$

#### Explanation:

Note that when $0 < t < 1$ then:

$\ln \left(1 - t\right) = - {\sum}_{n = 1}^{\infty} \left({t}^{n} / n\right) < - t$

So:

$\ln \left({\prod}_{n = 2}^{\infty} \left(2 - {2}^{\frac{1}{n}}\right)\right) = {\sum}_{n = 2}^{\infty} \ln \left(2 - {2}^{\frac{1}{n}}\right)$

$\textcolor{w h i t e}{\ln \left({\prod}_{n = 2}^{\infty} \left(2 - {2}^{\frac{1}{n}}\right)\right)} = {\sum}_{n = 2}^{\infty} \ln \left(1 - \left({2}^{\frac{1}{n}} - 1\right)\right)$

$\textcolor{w h i t e}{\ln \left({\prod}_{n = 2}^{\infty} \left(2 - {2}^{\frac{1}{n}}\right)\right)} \le - {\sum}_{n = 2}^{\infty} \left({2}^{\frac{1}{n}} - 1\right)$

Then:

2^(1/n) = e^(1/n ln 2) = sum_(k=0)^oo (1/n ln 2)^k/(k!) > 1 + 1/n ln 2

So:

${2}^{\frac{1}{n}} - 1 > \frac{1}{n} \ln 2$

So:

${\sum}_{n = 2}^{\infty} \left({2}^{\frac{1}{n}} - 1\right) \ge \ln 2 {\sum}_{n = 2}^{\infty} \frac{1}{n}$

diverges, since the harmonic series diverges.

So:

${\lim}_{N \to \infty} \ln \left({\prod}_{n = 2}^{N} \left(2 - {2}^{\frac{1}{n}}\right)\right) = - \infty$

and therefore:

${\prod}_{n = 2}^{\infty} \left(2 - {2}^{\frac{1}{n}}\right) = {\lim}_{t \to - \infty} {e}^{t} = 0$

$\textcolor{w h i t e}{}$
Footnote

For reference, here's a failed attempt that nevertheless illustrates a way you might approach such a problem...

I think it should be possible to prove that the product is $0$ using a method in a similar vein to the usual one for proving that the harmonic series diverges...

The basic idea is:

$\left(2 - {2}^{\frac{1}{2}}\right) \left(2 - {2}^{\frac{1}{3}}\right) \left(2 - {2}^{\frac{1}{4}}\right) \left(2 - {2}^{\frac{1}{5}}\right) \left(2 - {2}^{\frac{1}{6}}\right) \left(2 - {2}^{\frac{1}{7}}\right) \left(2 - {2}^{\frac{1}{8}}\right) \ldots$

$\le \underbrace{\left(2 - {2}^{\frac{1}{2}}\right)} \underbrace{\left(2 - {2}^{\frac{1}{4}}\right) \left(2 - {2}^{\frac{1}{4}}\right)} \underbrace{\left(2 - {2}^{\frac{1}{8}}\right) \left(2 - {2}^{\frac{1}{8}}\right) \left(2 - {2}^{\frac{1}{8}}\right) \left(2 - {2}^{\frac{1}{8}}\right)} \ldots$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{\le}}} \left(2 - {2}^{\frac{1}{2}}\right) \left(2 - {2}^{\frac{1}{2}}\right) \left(2 - {2}^{\frac{1}{2}}\right) \ldots = 0$

Hmmm...that does not quite work either, since:

$\left(2 - {2}^{\frac{1}{n}}\right) \left(2 - {2}^{\frac{1}{n}}\right) > \left(2 - {2}^{\frac{2}{n}}\right)$