7
$\begingroup$

Let $(a_n)_{n\geq 1}$ and $(b_n)_{n \geq 1}$ be two sequences of real numbers such that $$b_n=a_{n+2}-5a_{n+1}+6a_{n}, \: \forall n \geq 1$$ Prove that if $(b_n)$ is convergent, then $(a_n)$ is also convergent.

I defined $c_n=a_{n+1}-2a_n$ and the relation became $b_n=c_{n+1}-3c_n.$ Then I tried to prove that $c_n$ is convergent by expressing $c_n$ only in terms of $b_n, b_{n-1}, \dots b_1$ and $c_1$, but the convergence doesn't follow from here and I got stuck.

EDIT: As proven below, this statement is false !

$\endgroup$
5
$\begingroup$

Setting $b_n = 1$ (which is clearly convergent) and $a_1 = a_2 = 1$, we get $$ a_n = \frac16(3\cdot 2^n - 3^n + 3) $$ (WolframAlpha calculation), which doesn't converge.

$\endgroup$
4
$\begingroup$

Define $c_n=a_{n+1}-a_n$, then the recursion transforms to:

$$b_n=c_{n+1}-6c_n$$

If $c_n$ converges then $b_n$ converges. So, to disprove the statement it suffices to find a non-convergent sequence $a_n$ such that $a_{n+1}-a_n$ converges.

This is a classical problem. One such example is $a_{n}=\sqrt{n}$.

One can also find examples where $a_n$ is bounded and divergent.

$\endgroup$
3
$\begingroup$

This is false. Take for example $a_n=2^n$ (which is divergent) then $$b_n=a_{n+2}-5a_{n+1}+6a_{n}=2^{n+2}-5\cdot 2^{n+1}+6\cdot 2^{n} =(4-10+6)2^{n}=0$$ which is convergent.

$\endgroup$
2
$\begingroup$

This not true : if it were, if $b_n$ converged, you would have $c_n:=a_{n+1}-2a_n \rightarrow 0$.

But, just setting $c_n:= 3^n + \varepsilon_n$, you have $$ b_n= \varepsilon_{n+1}-3\varepsilon_n $$ you can take, then, for example, $\varepsilon_n:=\frac{1}{3^{n+1}}$, and you would have $b_n\rightarrow 0 $, $c_n\rightarrow +\infty$ and so, $(a_n)$ could not be convergent...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.