Determine for which value of m, the sequence (a_n ) is convergent and find the limit Let $a_1=1$ and $a_{n+1}=ma_n+10$ for $n\geq 2$. Determine for which value of m, the sequence $(a_n)$ is convergent and find the limit.
I know that if if we simplify we will get $0=L(M-1)+10$, which is equal to $10/(1-M)=L$ but I have no clue what to do from there
 A: If $m=1$, then $a_n=10(n-1)+1$ for all $n\in\mathbb N$, so the sequence does not converge. Suppose that $m\neq 1$. Then, we can prove that $a_{n+1}=m^n+10\frac{m^n-1}{m-1}$ for $n\geq 1$ by induction: this holds for $n=2$, and, if it holds for some $n\geq 2$, then $$a_{n+2}=ma_n+10=m\left(m^n+10\frac{m^n-1}{m-1}\right)+10=m^{n+1}+10\frac{m^{n+1}-m}{m-1}+10=$$$$m^{n+1}+10\left(\frac{m^{n+1}-m}{m-1}+1\right)=m^{n+1}+10\frac{m^{n+1}-1}{m-1}.$$ Therefore, this holds for all $n\in\mathbb N$, so $$a_{n+1}=m^n+10\frac{m^n-1}{m-1}=\frac{m^{n+1}+9m^n-10}{m-1}.$$ So, $a_n$ converges if and only if $m^{n+1}+9m^n=m^n(m+9)$ converges. So, either $m=-9$, or $m^n$ converges, which happens only if $|m|<1$. Hence $a_n$ converges only when $|m|<1$ or $m=-9$. In all of those case, you can check that the limit is $\frac{10}{1-m}$.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\color{#0000ff}{\large a_{n}}
&=
ma_{n - 1} + 10 = m^{2}a_{n - 2} + 10m + 10= m^{3}a_{n - 3} + 10m^{2} + 10m + 10
= \cdots
\\[3mm]&=
m^{n - 1}a_{1} + \sum_{k = 0}^{n - 2}10\,m^{k}
=
m^{n - 1} + 10\,{m^{n - 1} - 1\over m - 1}
=
\color{#0000ff}{\large{m^{n} + 9m^{n - 1} - 10 \over m - 1}\,,\qquad n \geq 1}
\end{align}
It converges whenever


*
*$m = -9$ since the factor $m^{9} + 9m^{n - 1}$ vanishes identically.

*Or $\verts{m} < 1$. In that case
$\lim_{n \to \infty} m^{n} = 0 = \lim_{n \to \infty} m^{n - 1}$.


Then, in both cases:
$$
\mbox{When}\quad \lim_{n \to \infty}\pars{m^{n} + 9m^{n - 1}} = 0\,,
\quad
\mbox{we get}\quad
\lim_{n \to \infty}a_{n} = {10 \over 1 - m}
$$
