# if $\lim_{n\to\infty}(4a_{n+2}-4a_{n+1}+a_{n})=2014$ prove the $\lim_{n\to\infty}a_{n}$ is exist and find the value

Let sequence $\{a_{n}\}$ such $$\lim_{n\to\infty}(4a_{n+2}-4a_{n+1}+a_{n})=2014$$

show that $$\lim_{n\to \infty}a_{n}$$ exist and find the limit value.

Now I use an ugly method to solve this. I use this follow lemma: if $$\lim_{n\to\infty}(a_{n+1}-\lambda a_{n})=a \Longleftrightarrow \lim_{n\to\infty}a_{n}=a,|\lambda|<1$$

I know this lemma proof is ugly, maybe anyone here has a simple method. Thank you.

Consider $b_0=a_0$, $b_1=a_1-a_0$, and $b_n=a_n-a_{n-1}+\frac14a_{n-2}$ for every $n\geqslant2$, and define the shift operator $\vartheta$ on sequences $c=(c_n)_{n\geqslant0}$ by $(\vartheta c)_n=c_{n-1}$ for every $n\geqslant1$ and $(\vartheta c)_0=0$.

Then the sequences $a=(a_n)_{n\geqslant0}$ and $(b_n)_{n\geqslant0}$ solve the identity $$b=Q(\vartheta)(a),\qquad Q(x)=1-x+\tfrac14x^2,$$ and the inverse of the polynomial $Q(x)$ in the field of formal series is $$Q(x)^{-1}=(1-\tfrac12x)^{-2}=\sum_{k\geqslant0}\frac{k+1}{2^k}x^k,$$ hence, $$a=Q(\vartheta)^{-1}(b)=\sum_{k\geqslant0}\frac{k+1}{2^k}\vartheta^k(b),$$ that is, for every $n\geqslant0$, $$a_n=\sum_{k=0}^n\frac{k+1}{2^k}b_{n-k}.$$ All this is algebra. Now, some analysis: the usual $N$-$\delta$ approach and the absolute convergence of the series $Q(x)^{-1}$ at $x=1$ show that, if $b_n\to\ell$ then $$a_n\to Q(1)^{-1}\ell=4\ell.$$ In the present case, one assumes that $b_n\to\frac14\cdot2014$ hence $a_n\to2014$.

Edit: The above is a bare-hands approach. If one can use the "ugly" lemma mentioned in the question, things are quicker. First, the lemma:

Assume that $a_{n+1}-\lambda a_{n}\to\ell$ for some $|\lambda|\lt1$, then $a_n\to\ell/(1-\lambda)$.

Now, its application in the case at hand: let $c_n=a_n-\frac12a_{n-1}$ and $b_n=c_n-\frac12c_{n-1}$, then $(b_n)$ is the sequence introduced in our answer above and $b_n\to\ell$ with $\ell=\frac14\cdot2014$ hence, using the lemma once with $\lambda=\frac12$, one gets $c_n\to\ell/(1-\frac12)=2\ell$, and using the lemma a second time, one gets $a_n\to(2\ell)/(1-\frac12)=4\ell$, QED.

Note finally that the proof of the "ugly" lemma is similar to the first approach in this post.

• Nice proof! Thank you+1 – china math Jun 22 '14 at 10:20