Find the limit of the sequences: $a_{n+1}=3a_n - n + 1$ and $(a_n)^\frac{1}{n}$ with $a_0 > 0 $ Let $a_0 > 0 $ and $$a_{n+1}=3a_n - n + 1.$$
I have to find its limit. I have also to find the limit of $(a_n)^\frac{1}{n}$. But this seems even more complicated. For the first part I've used the comparison $$a_n>\frac{n-1}{2}$$ and I've got that the limit is $\infty$.But I've no clue about the second limit. 
 A: You can use this way to do. Note
$$a_{n+1}=3a_n - n + 1, a_n=3a_{n-1}-n+2$$
and hence
$$ a_{n+1}-a_n=3(a_n-a_{n-1})-1.$$
or
$$ a_{n+1}-a_n-\frac12=3(a_n-a_{n-1}-\frac12).$$
So $a_{n}-a_{n-1}-\frac12=3^{n-1}(a_1-a_0-\frac12)$ or $a_{n}-a_{n-1}=\frac12+3^{n-1}(a_1-a_0-\frac12)$ and hence
\begin{eqnarray} 
a_n&=&a_0+(a_1-a_0)+(a_2-a_1)+\cdots+(a_n-a_{n-1})\\
&=&a_0+\sum_{k=1}^n[\frac{1}{2}+3^{k-1}(a_1-a_0-\frac12)]\\
&=&a_0+\frac{1}{4}[(3^n-1)(4a_0+1)+2n].
\end{eqnarray}
Thus $\lim_{n\to\infty}a_n^{1/n}=1$ if $4a_0+1=0$ and otherwise $\lim_{n\to\infty}a_n^{1/n}=3.$
A: You can find exactely what the solution to that reccurence is ($A\cdot3^n+\frac{1}{2}n-\frac{1}{4}$) and than just calculate the limit. 
A: Let's prove by induction that
$$a_n=3^na_0+b_n$$
where $n\le b_n\le3^{n-1}\quad(*)$.
This is certainly true for $a_0$. And
$$a_{n+1}=3a_n-n+1=3(3^na_0+b_n)-n+1=3^{n+1}a_0+3b_n-n+1$$
so let
$$b_{n+1}=3b_n-n+1$$
From $(*)$,
$$n+1\le 3n-n+1\le b_{n+1}\le3^{n}-n+1\le 3^n$$
which proves our assertion.
Now,
$$\lim_{n\to\infty}\sqrt[n]{3^n+n}\le\lim_{n\to\infty} \sqrt[n]{a_n}\le\lim_{n\to\infty}\sqrt[n]{3^n+3^{n-1}}$$
Thus, the limit is $3$.
