When does the 'if and only if' theorem hold for sequence limits?

[Jakim]

We know that for $k\in\mathbb{N}$ we have, if:

$\displaystyle\lim_{n\to\infty}\left(a_{n}-a_{n-k}\right)=k\cdot a$

then:

$\displaystyle\lim_{n\to\infty}\frac{a_n}{n}=a$

When the reverse impliaction is also true? What do we have to assume to achieve if and only if theorem? I'm especially interested in a case when $a=0$. I state (now without a proof, just intuition) that if

$\displaystyle\lim_{n\to\infty}\frac{a_n}{n}=a$

then $\displaystyle\limsup_{n\to\infty}\left(a_{n}-a_{n-k}\right)$ and $\displaystyle\liminf_{n\to\infty}\left(a_{n}-a_{n-k}\right)$ are bounded and if these limits are the same, they are equal $0$.

 

[jvicens]

I can confirm that the reverse implication is also true. In order for the "if and only if" theorem to hold, we have to assume that the sequence {a_n} is bounded. This means that there exists a real number M such that |a_n| ≤ M for all n ∈ ℕ.

In the case where a=0, we can say that the sequence {a_n} is bounded above by M and bounded below by -M. This means that the limit superior and limit inferior of the sequence {a_n} will also be bounded by M and -M respectively.

If the limit superior and limit inferior of the sequence {a_n} are equal and equal to 0, then we can conclude that the limit of the sequence {a_n} is also equal to 0. This is because the difference between a_n and a_n-k will approach 0 as n approaches infinity, as stated in the given condition.

In summary, in order for the "if and only if" theorem to hold, we have to assume that the sequence {a_n} is bounded. In the case where a=0, we can say that the limit superior and limit inferior of the sequence {a_n} are bounded and equal to 0, which implies that the limit of the sequence {a_n} is also equal to 0.