Question

Assume that the recursively defined sequence converges and find its limit
a1= 2; a (subscript) n+1= 48 / 2+ a(subscript)n.
What does this converge to? I had 24 but I think that's wrong? Is it 0 or 24?

Answer 1

Hold on. Let me clarify parentheses. So you have \(a_1=2\) and \[a_{n+1}=\frac{48}{2}+a_n?\]or\[a_{n+1}=\frac{48}{2+a_n}?\]

Answer 2

your second option, a(n+1) = [(48)/ (2+ a(subscript)n)]

Answer 3

If you're told to assume that it converges, this is actually rather easy. Just assume that you've reached a point where it's converged already. I.e. set \(a_{n+1}=a_n=x\) and solve for \(x\). So you just want to solve\[x=\frac{48}{2+x}\]for \(x\). You should get two solutions. Can you tell me what they are?

Answer 4

6, -8?

Answer 5

Right. So one of those values will be the value it converges to. To decide which one, simply notice that since you're starting at \(a_1=2\), which is a positive number, \(a_n\) will be positive for all \(n\in\mathbb{Z}^+\). So the limit can't possibly be \(-8\), and so it must converge to \(6\).

Did this make sense?

Answer 6

Hmm, It makes sense yes, are you sure its that easy?

Answer 7

If you're given that it converges, yes. Proving it converges is not quite so easy.

Answer 8

Alright, can i ask you one more brief question?

Answer 9

Sure.

Answer 10

I have to determine if the following is convergent or divergent, and i have it as convergent but im not sure..it is: sigma with infinity as top bound, k=3, and the notation itself is (3)/ (sq rt of k^2 -5)...

Answer 11

So\[\sum_{k=3}^\infty\frac{3}{\sqrt{k^2-5}}?\]

Answer 12

yes sir.

Answer 13

Well, since \(\sqrt{k^2-5}\) ~ \(k\), this is approximately \(\sum_{n=1}^\infty 1/n\) which diverges. So I would guess that it diverges. As for proving it... I might suggest trying a comparison test.

Answer 14

Of course. For \(k\ge3\) we have \(\sqrt{k^2-5}<k\), so\[\frac{3}{\sqrt{k^2-5}}>\frac{1}{\sqrt{k^2-5}}>\frac{1}{k}\]So\[\sum_{k=3}^\infty\frac{3}{\sqrt{k^2-5}}>\sum_{n=3}^\infty\frac{1}{n}=\infty\]

Answer 15

so it does indeed diverge alright, thank you

Answer 16

so anything that results in infinity like that diverges?

Answer 17

Yes. If it sums to infinity, people usually say it diverges.

Answer 18

Alright thanks a bunch.

Answer 19

No problem.