Need to show that this converges

let a1=1a_1=1 and a2=2a_2=2 and an+2=nan+1+ann+1a_{n+2}=\frac{na_{n+1}+a_n}{n+1}
I need to prove that this converges. Here is my try :

Let’s look at limn−>âˆ‍an+2\lim_{n->âˆ‍}a_{n+2} For an instance suppose it does converge and let \lim_{n->âˆ‍}a_{n+2}=L\lim_{n->âˆ‍}a_{n+2}=L

So L=\frac{nL+L}{n+1}L=\frac{nL+L}{n+1} and i know that n->âˆ‍n->âˆ‍ So right hand side of the equation will be L so L=LL=L is satisfied, therefore it does converge.

Would you say this way of thinking is correct? Or Should I try a formal one.

Btw, it’s given in question that L=1-e^{-1}L=1-e^{-1}, And I’m also asked to prove that L equals this.

I tried taking the lnln of each side, but couldn’t end anywhere.




You could try to prove that (a_{2n})(a_{2n}) and (a_{2n+1})(a_{2n+1}) adjacent.
– Abdallah Hammam
Oct 20 at 17:42



I’m sorry guys, I’ve wrote the values of a_1a_1 and a_2a_2 wrong. Please re-check 🙁
– Xenidia
Oct 20 at 17:43



suppose it does converge […] therefore it does converge That’s circular and proves nothing.
– dxiv
Oct 20 at 17:46



You will need to create a more formal proof. You assumed that the limit exists so you cannot use this to prove that the sequence converges; all you have done is not found any contradiction after assuming the limit exists.
– Hugh
Oct 20 at 17:47



Hint: a_{n+2}-a_{n+1}=-\frac{a_{n+1}-a_n}{n+1}a_{n+2}-a_{n+1}=-\frac{a_{n+1}-a_n}{n+1}
– Did
Oct 20 at 17:49


2 Answers


Sorry, but the argument is flawed under many respects.

Let’s try a different sequence, by way of example. Set

a_1=1,\qquad a_{n+1}=2a_n+1

a_1=1,\qquad a_{n+1}=2a_n+1

Suppose \lim_{n\to\infty}a_n=L\lim_{n\to\infty}a_n=L; then L=2L+1L=2L+1, so L=-1/2L=-1/2. Did we prove the sequence converges? Of course not! It would be absurd that a positive sequence converges to a negative limit.

Another example. Consider a_1=1a_1=1 and a_{n+1}=\frac{a_n+1}{4a_n+1}a_{n+1}=\frac{a_n+1}{4a_n+1}. If \lim_{n\to\infty}a_n=L\lim_{n\to\infty}a_n=L, then



so 4L^2+L=L+14L^2+L=L+1 that gives L=\pm\frac{1}{2}L=\pm\frac{1}{2}. However, the sequence is positive. What we have proved is that

if the sequence converges, then the limit is \frac{1}{2}\frac{1}{2}

But we have not proved convergence. This can be done in various ways, but it is not the main point.

Your argument is faulty also when you substitute a_na_n with LL. Both nn and a_na_n depend on nn and you cannot substitute one limit and not the other. It would be the same as saying that, since \lim_{x\to0}\sin x=0\lim_{x\to0}\sin x=0, then

\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{0}{x}=0

\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{0}{x}=0

which you surely know is utterly wrong. Anyway, if you find that L=LL=L you don’t even have a clue on what the limit can be. No information about convergence at all.



Great thanks. How do you think I should approach here? I see that the sequence is bounded from below, Should I look for whether its monotonic or not? Couldn’t think of any way
– Xenidia
Oct 20 at 18:08



@Xenidia Those are the standard attempts to try; maybe computing some terms can give you a clue about the thing. Also, since you’re given that the limit is 1-e^{-1}1-e^{-1}, you can try proving (by induction), that 11 is an upper bound.
– egreg
Oct 20 at 18:11


we have





a_{n+2}a_{n+2} is the barycenter of a_{n+1}a_{n+1} and a_na_n with respective weights nn and 11.

So all a_na_n are in [1,2][1,2].

You should prove by induction that

(a_{2n+1})(a_{2n+1}) is increasing while (a_{2n})(a_{2n}) is decreasing.

both are bounded.

So they converge and it’s not difficult to show that theire limits are equal.