Question

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

Answer 1

`prod_(k=2)^oo(2-root(k)(2)) = 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-root(n)(2)`, the condition `a_n < 1` is not sufficient for convergence to zero in

`lim_(n->oo)prod_(k=2)^n(2-2^(1/k))`

The necessary and sufficient condition for convergence of

`prod_(k=2)^oo a_k` to a non zero value is that `sum_(k=2)^oo log(a_k)` converges.

If `a_n < 1` and `sum_(k=2)^oo log(a_k)` does not converge, then
`prod_(k=2)^oo a_n` converges to `0`.

Now considering the inequality `(x-1)/x le log(x), x > 1` we have

`(1 - 2^(1/k))/(2 - 2^(1/k)) le log(2-2^(1/k))`

Considering now

`sum_(k=2)^oo (1 - 2^(1/k))/(2 - 2^(1/k))`

with `b_k = (1 - 2^(1/k))/(2 - 2^(1/k))` clearly this series does not converge because `b_k > 0` and

`lim_(k->oo)b_k/b_(k+1)=1`

so because `sum_(k=2)^oo (1 - 2^(1/k))/(2 - 2^(1/k)) le sum_(k=2)^oo log(2-2^(1/k))`

`sum_(k=2)^oo log(2-2^(1/k))` does not converge then finally

`lim_(n->oo)prod_(k=2)^n(2-2^(1/k)) = 0`

Answer 2

`prod_(n=2)^oo(2-2^(1/n)) = 0`

Explanation:

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

`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`

`color(white)()`
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:

`(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))`