For the answer of this question, I am told that it is $a > 0$, but if $a$ is $1$ then would the resulting fraction not be $\frac{0}{2}$. Is it still converging then?
Am I wrong in thinking the answer should be $a > 1$ instead?
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Sign up to join this communityFor the answer of this question, I am told that it is $a > 0$, but if $a$ is $1$ then would the resulting fraction not be $\frac{0}{2}$. Is it still converging then?
Am I wrong in thinking the answer should be $a > 1$ instead?
There is nothing wrong with a feaction like $0/2$ where just the numerator is zero; such a fraction is just zero. The problem is if the _ denominator_ is zero.
As commented, you have
$$u_n=\left(\frac{a-1}{a+1}\right)^nu_0$$
and the above converges (to a finite value, that is) iff ( since you're given $\;a>0\;$)
$$\left|\frac{a-1}{a+1}\right|\le1\iff |a-1|\le|a+1|= a+1$$
Now, we can do:
$$\begin{align*}&0<a< 1: \implies -a+1\le a+1\implies 2a\ge0\;,\;\;\text{and this is true in any such case}\\{}\\ &a\ge1: \implies a-1\le a+1\iff -1\le 1\,,\,\,\text{and again this is true in any such case}\end{align*}$$
Thus, the sequence $\;\{u_n\}\;$ converges to a finite value, which is either $\;u_0\;$ o $\;0\;$ , for any value of $\;a>0\;$ .
Hint for a geometric solution:
As a geometric sequence, it converges if and only if $\;\dfrac{\lvert a-1\rvert}{\lvert a+1\rvert}\iff\lvert a-1\rvert<\lvert a+1\rvert$.
Now interpret in terms of distance: if $\;b>c$, $$\lvert a-b\rvert <\lvert a-c\rvert\iff a > \frac{b+c}2.$$