1. ## sqrt(2)

Show that

sqrt(2), sqrt(2*sqrt(2)), sqrt(2*sqrt(2*sqrt(2))), ...

converges, and find the limit.

--------------------------

So, uh, to show that it converges, we can use the monotone convergence theorem, that says if its bounded and monotone, it will converge... maybe?

EDIT:

Oh, maybe use the theorem that says that subsequences of a convergent sequence converge to the same lim as the orig. sequence? I'm not too sure.

2. $a_1=2^{\frac{1}{2}}$
$a_2=2^{\frac{3}{4}}$
$a_3=2^{\frac{7}{8}}$.
By induction, $a_n=2^{\frac{2^n-1}{2^n}}$.
Then $\lim_{n\to\infty}a_n=2$, so $a_n$ is convergent.

3. Originally Posted by seerTneerGevoLI
Show that

sqrt(2), sqrt(2*sqrt(2)), sqrt(2*sqrt(2*sqrt(2))), ...

converges, and find the limit.
.
Hint: $a_1 = \sqrt{2}$ and $a_{n+1} = \sqrt{2a_n}$.

4. Originally Posted by red_dog
$a_1=2^{\frac{1}{2}}$
$a_2=2^{\frac{3}{4}}$
$a_3=2^{\frac{7}{8}}$.
By induction, $a_n=2^{\frac{2^n-1}{2^n}}$.
Then $\lim_{n\to\infty}a_n=2$, so $a_n$ is convergent.
So red_dog pretty much nailed it. I'll break it down for you.

We have the sequence $x_n= \sqrt{2x_{n-1}} , x_1= \sqrt{2}$ which is strictly increasing and bounded (this would require a rigorous proof...but since you just said show, I'll show it in a simple way).

We have:

$
x_1^2 = 2
\frac{ x_2^4}{4} = 2
\frac{ x_3^8}{64} = 2
$

See the pattern?

$
x_3^8 = 2^7
$

In general, $x_n^{2^n} = 2^{2^n -1}$

Use the log rule: $\log_b y^k = k \log_b y$

We have: $2^n \log_2 x_n = 2^n -1$. Divide both sides by $2^n$ and thus: $x_n = 2^{ \frac{2^n -1}{2^n}}$.

Take the limit as n goes to infinity:

$\lim_{n\to \infty} x_n = 2$

5. Originally Posted by AfterShock

We have: $2^n \log_2 x_n = 2^n -1$. Divide both sides by $2^n$ and thus: $x_n = 2^{ \frac{2^n -1}{2^n}}$.

Take the limit as n goes to infinity:

$\lim_{n\to \infty} x_n = 2$
How do you get that the lim is equal to 2 from $x_n = 2^{ \frac{2^n -1}{2^n}}$...

6. $\lim _{n \to \infty } \left( {\frac{{2^n - 1}}{{2^n }}} \right) = \lim _{n \to \infty } \left( {1 - \frac{1}{{2^n }}} \right) = 1\;\& \;2^1 = 2$

7. Originally Posted by Plato
$\lim _{n \to \infty } \left( {\frac{{2^n - 1}}{{2^n }}} \right) = \lim _{n \to \infty } \left( {1 - \frac{1}{{2^n }}} \right) = 1\;\& \;2^1 = 2$
Hmm, I discussed this with my professor, and he said something along the lines of there being no theorem stating something like:

lim a^(c_n) = lim a^(lim c_n)

EDIT:

You're doing 2^(lim c_n) ...

8. Originally Posted by seerTneerGevoLI
Hmm, I discussed this with my professor, and he said something along the lines of there being no theorem stating something like:

lim a^(c_n) = lim a^(lim c_n)

EDIT:

You're doing 2^(lim c_n) ...
The function $f(x) = a^x$ for $a>0$ is a continous function. So if $c_n$ is a convergent sequence with limit $c$ then $a^{c_n}$ is a convergent sequence with limit $a^c$ by definition.