1) let x_1 > 1 and x_(n+1) := (2-1) / x_n for n c X. Show that x_n is decreasing and bounded below by 2. find the limit.
2) let x_n := 1/(1^2) + 1/(2^2) + ... + 1/(n^2) for each n c N. Prove that x_n is increasing and bounded and hence converges. [Hint: note that if k>= 2, then 1/(k^2) =< 1/(k(k-1)) = 1/(k-1) - 1/k]
THANK YOU!
Focus, I applaud your effort in this thread.
However, please review this other thread.
It has been my considerable experience that when someone is so wanting in being able to post in correct notation, that person will not understand any reply given.
I apologize for my intervention but the first proposed 'quiz' is of remarkable interest... also because it has do be 'decrypted' ...
Under the hypothesis that is ...
, (1)
... the solution is 'trivial' ...
(2)
... and the clearly doesn't exist...
However under the hypothesis that is...
, (2)
... the proposed problem is very nice! ... The first terms of the sequence in this case are ...
(4)
The expressions that are in (4) are called 'continued fractions' and they interpret one of the most 'sggestive' chapter of Math. If we denote with the , at first finding seems to be an 'hard' eneterprise. By observing with care (4) however You 'discover' that is...
(5)
... the only 'valid' solution of which is . Therefore is...
(5)
... no matter which is ...
Kind regards
Osumath dude. I am taking 547 this quarter as well I have completed 1...and am working on 2! Let me know what you find!
Here is my 1:
Lemma: If a>1 then 2-(1/a)>1.
Proof 0<1/a<1...then -1/a>-1 2+(-1/a)>2-1=1. by induction xsubn>1 for all n in N.
We wish to show if a>1 then 2-1/a<a. Which will be a polynomial, and therefore 1<a and 0<(a-1)^2 is always greater than 0.
Now we know that xsubn is bounded and monotone.
To find the limit solve for L in L+1/L=2.
Anyone how does this abbreviated proof sound????