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

##### 2 Answers

#### Explanation:

At first glance, we would propose

So calling

The necessary and sufficient condition for convergence of

If

Now considering the inequality

Considering now

with

so because

#### Explanation:

Note that when

#ln(1-t) = -sum_(n=1)^oo (t^n/n) < -t#

So:

#ln (prod_(n=2)^oo(2-2^(1/n))) =sum_(n=2)^oo ln(2-2^(1/n))#

#color(white)(ln (prod_(n=2)^oo(2-2^(1/n))))=sum_(n=2)^oo ln(1 - (2^(1/n) - 1))#

#color(white)(ln (prod_(n=2)^oo(2-2^(1/n))))<= -sum_(n=2)^oo (2^(1/n) - 1)#

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^(1/n)-1 > 1/n ln 2#

So:

#sum_(n=2)^oo (2^(1/n)-1) >= ln 2 sum_(n=2)^oo 1/n#

diverges, since the harmonic series diverges.

So:

#lim_(N->oo) ln(prod_(n=2)^N(2-2^(1/n))) = -oo#

and therefore:

#prod_(n=2)^oo(2-2^(1/n)) = lim_(t->-oo) e^t = 0#

**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

The basic idea is:

#(2-2^(1/2))(2-2^(1/3))(2-2^(1/4))(2-2^(1/5))(2-2^(1/6))(2-2^(1/7))(2-2^(1/8))...#

#<= underbrace((2-2^(1/2)))underbrace((2-2^(1/4))(2-2^(1/4)))underbrace((2-2^(1/8))(2-2^(1/8))(2-2^(1/8))(2-2^(1/8)))...#

#color(red)(cancel(color(black)(<=))) (2-2^(1/2))(2-2^(1/2))(2-2^(1/2))... = 0#

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

#(2-2^(1/n))(2-2^(1/n)) > (2-2^(2/n))#