1. Converge / Diverge

Which ones are convergent and which are divergent and what are the limits of the convergent ones?
a) lim (as n goes to infinity) of (-1)^n sin(n)/n

b) lim (as n goes to infinity) of (-1)^n sin(1/n)

c) lim (as n goes to infinity) of n^(1/n) (L'Hospital's Rule)

d) lim (as n goes to infinity) a(sub)n where a1 = sqrt(2) , a2 = sqrt(2sqrt(2)), a3 = sqrt(2sqrt(2sqrt2))) , an+1 = sqrt(2an)

i think a is divergent becaue of the alternator and b converges to 0 because of squeeze theorem. Is this right? And how do i find the other two?

2. For a, look at the $\displaystyle\frac{sin(n)}{n}$. What happens when n goes to infinity there?

For b, concentrate on the sin portion.

For c, set equal to y and natural log.

For d, I think this should work $\displaystyle x=\sqrt{2\sqrt{2\sqrt{2....}}}$

3. so the first two converge to 0?
lny = (1/n) ln (n)? it says to use L'Hospital's Rule for c
im not sure what you mean for d, but would it just diverge?

4. First 2 are 0.

Ok so now we have, for the 3rd, $\displaystyle \lim_{n\to\infty}\frac{ln(n)}{n}=\frac{\infty}{\in fty}$ So now use L'Hopital's Rule. Remember to exponentiate the answer to undo the natural log.

However, you should know this goes to zero since y=x grows faster than y=ln(x).

$\displaystyle x=\sqrt{2\sqrt{2\sqrt{2....}}}\rightarrow x^2=2\sqrt{2\sqrt{2\sqrt{2....}}} \ \mbox{but x=\sqrt{2\sqrt{2\sqrt{2....}}}} \ x^2=2x$

5. Originally Posted by Playthious
Which ones are convergent and which are divergent and what are the limits of the convergent ones?
a) lim (as n goes to infinity) of (-1)^n sin(n)/n

b) lim (as n goes to infinity) of (-1)^n sin(1/n)

c) lim (as n goes to infinity) of n^(1/n) (L'Hospital's Rule)

d) lim (as n goes to infinity) a(sub)n where a1 = sqrt(2) , a2 = sqrt(2sqrt(2)), a3 = sqrt(2sqrt(2sqrt2))) , an+1 = sqrt(2an)

i think a is divergent becaue of the alternator and b converges to 0 because of squeeze theorem. Is this right? And how do i find the other two?
The solution of d) is fully elementar ifn someone follows the procedure I illustrated a lot of times. The 'recursive' equation can be written as...

$\displaystyle \Delta_{n} = a_{n-1}-a_{n} = \sqrt{2 a_{n}} - a_{n} = f(a_{n})$ (1)

The function $f(x)= \sqrt{2 x} - x$ is represented here...

There is only one 'attractive fixed point' in $x_{0}=2$ and because is $|f(x)|< |x_{0}-x|$ [see 'red line'...] any $a_{0}>0$ will generated a sequence monotonically convergent at $x_{0}$...

Kind regards

$\chi$ $\sigma$