3
$\begingroup$

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 $f_n(x) \rightarrow 0$ ?

$\endgroup$

3 Answers 3

1
$\begingroup$

Let $a_n=(n+1)c^n$ and note that by ratio test

$$\frac{a_{n+1}}{a_n}=\frac{(n+2)c^{n+1}}{(n+1)c^n}=\frac{(n+2)}{(n+1)}\cdot c\to 1\cdot c=c<1$$

thus

$$(n+1)(1-x)c^n \to 0$$

$\endgroup$
1
$\begingroup$

Assume $c > 0$ then $c^n = e^{nlnc}$ and since $c < 1 \ lnc < 0 $ then $c^n \sim e^{-n}$ but $\frac{n^k}{e^n} \rightarrow 0, \forall \ k$

$\endgroup$
1
$\begingroup$

$a_n =( n+1)c^n$, 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?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .