Sequence Divergence and Convergence Questions

[ardentmed]

Hey guys, I have a couple more questions.

View attachment 2801

For the first one, taking the limit to infinity obviously equals 0 so it should be convergent, right?

Also, for the second one, the limit as n approaches infinity for gives me indeterminate form, so I took the derivative which just gave me ln(n) which is 0 (and thus convergent). Although I'm really not sure about this one.

How would i go about proving the third one, aside from a1 < a2 for convergence? Would I just substitute values into prove that √2 < √(2+√2)) Can I get some help on this one?

Thanks in advance.

 

[Prove It]

Hey guys, I have a couple more questions.

View attachment 2801

For the first one, taking the limit to infinity obviously equals 0 so it should be convergent, right?

Also, for the second one, the limit as n approaches infinity for gives me indeterminate form, so I took the derivative which just gave me ln(n) which is 0 (and thus convergent). Although I'm really not sure about this one.

How would i go about proving the third one, aside from a1 < a2 for convergence? Would I just substitute values into prove that √2 < √(2+√2)) Can I get some help on this one?

Thanks in advance.

Even though it oscillates, $\displaystyle \begin{align*} \frac{ (-1)^n }{n} \end{align*}$ still approaches 0 as $\displaystyle \begin{align*} n \to \infty \end{align*}$, so the second sequence is still convergent.

As for the third, notice that by the definition of your sequence, you have $\displaystyle \begin{align*} a_n = \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{ 2 + \sqrt{\dots } } } } } \end{align*}$, with n of these roots. So as $\displaystyle \begin{align*} n \to \infty \end{align*}$ we have these roots going on forever.

Now notice that underneath the first square root, because these roots go on forever, where the second root starts is really the same thing as what you are trying to find, so you should find

$\displaystyle \begin{align*} a_n &= \sqrt{2 + a_n } \\ \left( a_n \right) ^2 &= 2 + a_n \\ \left( a_n \right) ^2 - a_n - 2 &= 0 \\ \left( a_n - 2 \right) \left( a_n + 1 \right) &= 0 \\ a_n - 2 &= 0 \textrm{ or } a_n + 1 = 0 \\ a_n &= 2 \textrm{ or }a_n = -1 \end{align*}$

Clearly since you start with a positive amount and keep adding to it, you must get a positive value, thus the sequence converges to 2.

 

[chisigma]

Hey guys, I have a couple more questions.

View attachment 2801

For the first one, taking the limit to infinity obviously equals 0 so it should be convergent, right?

Also, for the second one, the limit as n approaches infinity for gives me indeterminate form, so I took the derivative which just gave me ln(n) which is 0 (and thus convergent). Although I'm really not sure about this one.

How would i go about proving the third one, aside from a1 < a2 for convergence? Would I just substitute values into prove that √2 < √(2+√2)) Can I get some help on this one?

Thanks in advance.

The question c) is the most interesting. You have the4 difference equation...

$\displaystyle a_{n+1} = \sqrt{2 + a_{n}}\ (1)$

... that can be written as...

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

The function f(x) is illustrated here... View attachment 2807

... and it has only an attractive fixed point in x=2 [in x=-1 is $f(x) \ne 0$...] and because for $x \ge -2$ is $\frac{|f(x)|}{|x-2|} < 1$ any initial value $a_{0} \ge -2$ will produce a sequence monotonically convergent to 2...

Kind regards

$\chi$ $\sigma$