thiago_j
Sep24-04, 08:38 AM
Guys,
I'm trying to prove by induction that the sequence given by a_{n+1}=3-\frac{1}{a_n} \qquad a_1=1 is increasing and a_n < 3 \qquad \forall n .
Is the following correct? Thank you. :smile:
Task #1.
n = 1 \Longrightarrow a_2=2>a_1 is true.
We assume n = k is true. Then,
3-\frac{1}{a_{k+1}} > 3-\frac{1}{a_k}
a_{k+2} > a_{k+1} is true for n=k+1 .
This shows, by mathematical induction, that
a_{n+1} > a_{n} \qquad \forall n .
Task #2
We already know that
a_1 < 3 is true.
We assume n=k is true. Then,
a_k < 3
\frac{1}{a_k} > \frac{1}{3}
-\frac{1}{a_k} < -\frac{1}{3}
3-\frac{1}{a_k} < 3-\frac{1}{3}
a_{k+1} < \frac{8}{3} < 3
a_{k+1} < 3 is true for n = k+1 . Thus,
a_{n} < 3 \qquad \forall n .
I'm trying to prove by induction that the sequence given by a_{n+1}=3-\frac{1}{a_n} \qquad a_1=1 is increasing and a_n < 3 \qquad \forall n .
Is the following correct? Thank you. :smile:
Task #1.
n = 1 \Longrightarrow a_2=2>a_1 is true.
We assume n = k is true. Then,
3-\frac{1}{a_{k+1}} > 3-\frac{1}{a_k}
a_{k+2} > a_{k+1} is true for n=k+1 .
This shows, by mathematical induction, that
a_{n+1} > a_{n} \qquad \forall n .
Task #2
We already know that
a_1 < 3 is true.
We assume n=k is true. Then,
a_k < 3
\frac{1}{a_k} > \frac{1}{3}
-\frac{1}{a_k} < -\frac{1}{3}
3-\frac{1}{a_k} < 3-\frac{1}{3}
a_{k+1} < \frac{8}{3} < 3
a_{k+1} < 3 is true for n = k+1 . Thus,
a_{n} < 3 \qquad \forall n .