I'm trying to find the radius of convergence for
$(x-2)^n/(2x+1)
I did the ratio test and ended up with:
absolute value[(x-2)/(2x+1)]<1
How would I solve for the inequality at this point?
If |(x-2)/(2x+1)| < 1 then
(x-2)^2 < (2x+1)^2
x^2 - 4x + 4 < 4x^2 + 4x + 1
3x^2 + 8x - 3 > 0
x < -3 or x > 1/3
To solve the inequality $|\frac{x-2}{2x+1}| < 1$, you can follow these steps:
Step 1: Remove the absolute value signs by considering two cases:
Case 1: $\frac{x-2}{2x+1} > -1$
In this case, you can simply ignore the absolute value sign and write the inequality as $\frac{x-2}{2x+1} < 1$.
Case 2: $\frac{x-2}{2x+1} < 1$
In this case, you need to flip the inequality sign and write the inequality as $\frac{x-2}{2x+1}> -1$.
Step 2: Solve each case separately:
Case 1: $\frac{x-2}{2x+1} < 1$
Multiply both sides of the inequality by $2x + 1$:
$(x - 2) < (2x + 1)$
Expand the brackets:
$x - 2 < 2x + 1$
Rearrange the equation:
$x - 2x < 1 + 2$
Simplify:
$-x < 3$
Divide by -1 (which flips the inequality sign):
$x > -3$
So, for this case, the solution is $x > -3$.
Case 2: $\frac{x-2}{2x+1} > -1$
Multiply both sides of the inequality by $2x + 1$:
$(x - 2) > -(2x + 1)$
Expand the brackets:
$x - 2 > -2x - 1$
Rearrange the equation:
$x + 2x > -1 + 2$
Simplify:
$3x > 1$
Divide by 3:
$x > \frac{1}{3}$
So, for this case, the solution is $x > \frac{1}{3}$.
Step 3: Combine the solutions from both cases:
Since the radius of convergence represents the range of values for which the series converges, we need to consider the intersection of the solutions from both cases.
Taking the intersection of $x > -3$ and $x > \frac{1}{3}$, we find that the final solution is $x > \frac{1}{3}$.
Therefore, the radius of convergence for the given series is $x > \frac{1}{3}$.