<s>[INST] Why $ (n+1)(1-x)c^n \rightarrow_{n \rightarrow +\infty} 0$ Simple question.
Why the function $f_n(x) = (n+1)(1-x)c^n \rightarrow_{n \rightarrow +\infty} 0$ if $|c| < 1$.
I know that $(1-x)c^n \rightarrow 0$ but $(n+1) \rightarrow + \infty$. Why $f_n(x) \rightarrow 0$ ?
 [/INST] $a_n =( n+1)c^n$, where $|c| < 1.$
$|a_n| = (n+1)(|c|^n)$ .
Set $b:=1/|c|$, where $b >1$.
$b = 1+a$, $a>0$.
$b^n =(1+a)^n =$
$1+ na +n(n-1)/2! + ...$
Hence:
$|a_n| = \dfrac{n+1}{b^n}=$
$\dfrac{n+1}{1+na+(n(n-1)/2!)a^2 +...} \lt$
$\dfrac{2(n+1)}{n(n-1)a^2}=$
$\dfrac{2}{(n-1)a^2} +\dfrac{2}{n(n-1)a^2}.$
The limit $n \rightarrow \infty $ is?
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