How to solve convergences when dealing with series

[Trail_Builder]

Homework Statement

shows that the sequence $$U(n+1) = \sqrt{2U(n) + 5}$$ where U(1) = 3 converges to a limit, u, and find the value of u correct to 2 decimal places.

Homework Equations

The Attempt at a Solution

I only know how to solve convergences when dealing with series lol.

Any help please :)

thanks

 

[Dick]

If the sequence does have a limit L, then the limit must satisfy L=sqrt(2L+5). Do you see why? As for the proof part can you show the sequence is increasing and bounded?

 

[Trail_Builder]

ooooo

yeah I see why L=sqrt(2L+5).

I still can't go anywhere from here though :S

 

[Dick]

Ok. Take the function f(x)=sqrt(2x+5). You get the elements in the sequence by finding f(3), f(f(3)), f(f(f(3))) etc. To show that f(x)>x (that the sequence is increasing), you want to show f(x)-x>0. For what values of x is this true?

 

[HallsofIvy]

ooooo

yeah I see why L=sqrt(2L+5).

I still can't go anywhere from here though :S

You can't solve that for L? L²= 2L+ 5 so L²- 2L= 5. You can solve that by completing the square: L²- 2L+ 1= 5+ 1= 6. (L- 1)²= 6 so $L- 1= \pm\sqrt{6}$ and $L=1\pm\sqrt{6}$. Here, since the sequence starts at 3 and increases, $L= 1+ \sqrt{6}$. Of course, we need to prove that the sequnce is increasing and bounded above so it will converge.

Prove U(n)< U(n+1) for all n, by induction. If n= 1, then U(1)= 3 while U(2)= $2\sqrt{2U(1)+5}= \sqrt{6+ 5}=\sqrt{11}> 3$ (because 11> 9).
Now, suppose for some k, U(k+1)> U(k). Then U((k+1)+ 1)= $\sqrt{2U(k+1)+ 5}> \sqrt{2U(k)+ 5}= U(k+1)$.

You will also need to prove that the sequence is bounded above. Since (if the sequence has a limit) the limit is $1+ \sqrt{6}$ which is approximately 3.5so you might try proving, again by induction, that U(n)< 4 for all n.

 

[Trail_Builder]

o sick

thanks for help guys :D

much appreciated

I see it now :)