If $b_n$ is convergent, then $a_n$ is also convergent

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.


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.


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.

Tags:

Limits

Sequences And Series

Recurrence Relations

Real Analysis

Convergence Divergence

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