+50 Please, can someone just understand me what I'm doing wrong in the following attempt, without proposing other solutions? Thanks!
Problem: suppose $0
My attempt is the following: since $0
Since $6(1+a_n)$ is positive too, multiplying by $6(1+a_n)$ in $\frac{1}{9} < \frac{1}{7+a_n} < \frac{1}{7}$ leads to $\frac{2}{3}(1+a_n) < 6\frac{1+a_n}{7+a_n}<\frac{6}{7}(1+a_n)$.
But as said before, it is $1<1+a_n<3$, so $\frac{2}{3} <\frac{2}{3}(1+a_n) < 6\frac{1+a_n}{7+a_n}<\frac{6}{7}(1+a_n) < \frac{18}{7}$; of course I am wrong, because the function $f(x)=6\frac{1+x}{7+x}$ has indeed maximum $2$ and minimum $\frac{6}{7}$.
I don't understand where I am making mistake, probably I didn't flip correctly an inequality, but I can't see where. Thank you!
+129845 If I understand this...let an = x
1/9 < 1 /( 7 + x) < 1/7
1/9 < 1 /( 7 + x) 1 / (7 + x) < 1/7
7 + x < 9 7 < 7 + x
x < 2 x > 0
0 < x < 2
If we want an integer soliution x = an = 1
+129845 6/7 < 6 ( 1 + x) / (7 + x) < 2
6/7 < 6 (1 + x) / ( 7 + x) divide through by 6
1/7 < (1 + x) / (7 + x)
1 ( 7 +x) < 7 (1 + x)
7 + x < 7 + 7x
0 < 6x
0 < x
And
6 ( 1 + x) / (7 + x) < 2 divide through by 6
(1 + x) /( 7 + x) < 1/3
3 ( 1 + x) < 1 (7 + x)
3 + 3x < 7 + x
2x < 4
x < 2
0 < x < 2
Which is what we assumed
