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

Question

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

Isaiah Allen · Accepted Answer

#prod_(k=2)^oo(2-root(k)(2)) = 0#

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#

Luke Alvarez · Answer

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

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))#

Luke Alvarez · Answer

undefined To solve the limit $\lim_{n \to \infty} (2 - \sqrt{2})(2 - \sqrt[3]{2})(2 - \sqrt[4]{2}) \ldots (2 - \sqrt[n]{2})$, we recognize that each term in the product is of the form $2 - \sqrt[k]{2}$, where $k$ varies from $2$ to $n$.

Now, let's rewrite each term in a way that allows us to use a limit property. We can express $2 - \sqrt[k]{2}$ as $\sqrt[k]{2} - 2$. Then, we can rewrite the expression as $\lim_{n \to \infty} (\sqrt[k]{2} - 2)$ as $k$ ranges from $2$ to $n$.

Next, we notice that as $k$ approaches infinity, $\sqrt[k]{2}$ approaches $1$. Therefore, $\sqrt[k]{2} - 2$ approaches $1 - 2 = -1$.

Since the limit of each term in the product approaches $-1$, and we have an infinite number of such terms, the overall limit of the product is $(-1)^{\infty}$.

However, $(-1)^{\infty}$ is an indeterminate form. To resolve this, we can apply the natural logarithm and rewrite the expression as $\lim_{n \to \infty} \ln[(2 - \sqrt{2})(2 - \sqrt[3]{2})(2 - \sqrt[4]{2}) \ldots (2 - \sqrt[n]{2})]$.

Using the property $\ln(ab) = \ln(a) + \ln(b)$, we can rewrite the expression as $\lim_{n \to \infty} [\ln(2 - \sqrt{2}) + \ln(2 - \sqrt[3]{2}) + \ln(2 - \sqrt[4]{2}) + \ldots + \ln(2 - \sqrt[n]{2})]$.

As $n$ approaches infinity, each term in the sum approaches $\ln(2 - 1) = \ln(1) = 0$.

Therefore, the overall limit is $0$.

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

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