The skill here is finding the interval of convergence for a power series. To determine the radius, the ratio test gives lim |x-4| $n^2$ / $(n+1)^2$ = |x-4| < 1, so radius R=1 and open interval 3 < x < 5. Check endpoint x=3: sum $(-1)^n$ / $n^2$ converges absolutely by p-series with p=2 >1. At x=5: sum $1/n^2$ also converges by p-series. A tempting distractor is (3,5), failing because it excludes the endpoints where the series actually converges. A transferable convergence-interval strategy is to first compute the radius using the ratio or root test, then evaluate convergence at each endpoint using tests like p-series, alternating series, or comparison.

The skill here is finding the interval of convergence for a power series. This is a geometric series with ratio $r = \frac{2x-1}{7}$, converging when $|r| < 1$, so $|2x-1| < 7$, yielding interval $-3 < x < 4$ and radius $\frac{7}{2}$ centered at $x=\frac{1}{2}$. At endpoints $x=-3$ and $x=4$, $|r|=1$, so the geometric series diverges. No convergence at endpoints. A tempting distractor is $[-3,4]$, failing because geometric series diverge at $|r|=1$. A transferable convergence-interval strategy is to first compute the radius using the ratio or root test, then evaluate convergence at each endpoint using tests like p-series, alternating series, or comparison.

The skill here is finding the interval of convergence for a power series. To determine the radius, the ratio test yields $\lim |x-3| \frac{n}{n+1} = |x-3| < 1$, so R=1 and open interval $2 < x < 4$. Check endpoint x=2: $\sum(-1)^n / n$ converges by alternating series test. At x=4: $\sum 1/n$ diverges as harmonic series. A tempting distractor is $(2,4)$, failing because it excludes x=2 where the series converges. A transferable convergence-interval strategy is to first compute the radius using the ratio or root test, then evaluate convergence at each endpoint using tests like p-series, alternating series, or comparison.

The skill being tested here is finding the radius of convergence for a power series. Apply the ratio test to $(x-4)^{2n}$$/5^n$, giving a limit of $(x-4)^2$/5, which is less than 1 when |x-4| < √5. This indicates the radius is √5. The even powers effectively make it a geometric series in terms of $(x-4)^2$/5. A tempting distractor like R=1/√5 fails by incorrectly applying the root test without accounting for the exponent. Always remember to use the ratio or root test for the radius and separately test endpoints for conditional convergence in power series problems.

The skill being tested here is finding the radius of convergence for a power series. Apply the ratio test to the coefficients $(2n)!/(n!)^2$ $x^n$, resulting in a limit of 4|x| as n approaches infinity. The series converges when 4|x| < 1, so |x| < 1/4, giving a radius of 1/4. This central binomial coefficient series is known to have this radius, confirming the calculation. A tempting distractor like R=1/2 fails because it miscalculates the limit by overlooking the factorial growth rates properly. Always remember to use the ratio or root test for the radius and separately test endpoints for conditional convergence in power series problems.

The skill being tested is finding the radius of convergence for a power series. Term $\frac{(x-3)^n}{n (-2)^n}$, ratio $\frac{|x-3|}{|-2|} \cdot \lim \frac{n}{n+1} = \frac{|x-3|}{2} < 1$, R=2. The negative in denominator doesn't affect radius. No endpoints. Tempting R=-2 invalid as radius positive. Include absolute value in ratio for correct R.

The skill here is finding the radius and interval of convergence for a power series. To find the radius, apply the ratio test: $ \lim_{n \to \infty} \frac{ \left| \frac{(x-2)^{n+1}}{(n+1) (n+1)^2} \right| }{ \left| \frac{(x-2)^n}{n n^2} \right| } = |x-2| \cdot \frac{n^3}{(n+1)^3} \to |x-2| < 1 $, so $R=1$. Since the question asks only for the radius, we stop here. The n^3 in the denominator gives $R=1$. A tempting distractor is $R=\infty$, but this fails because the polynomial growth does not outpace the exponential. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.

The skill being tested is finding the radius of convergence for a power series. $ \sum(x+1)^n / 7^n $, geometric, $ |x+1|/7 < 1 $, $ R=7 $. Simple ratio. No endpoints. Tempting $ R=1/7 $ inverts. Apply geometric series formula directly.

The skill here is finding the interval of convergence for a power series. Recognize this as a geometric series $\sum[(x-4)/4]^n$, which converges when $|(x-4)/4| < 1$, so $|x-4| < 4$, giving open interval $(0, 8)$. Then, check the endpoints: at x=0, r=-1, $\sum(-1)^n$ diverges; at x=8, r=1, $\sum 1^n$ diverges. Neither endpoint converges. A tempting distractor is [0,8], but this fails because geometric series diverge at $|r|=1$. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.

The skill being tested is finding the interval of convergence for a power series. $\sum \frac{(x-2)^n}{3^n}$, geometric with $r=\frac{x-2}{3}$, $|r|<1$, $|x-2|<3$, $(-1,5)$. Endpoints: at -1, $\sum \left[ \frac{-3}{3} \right]^n = \sum(-1)^n$, alternates but $|$term$|=1$ not to 0, diverges. At 5, $\sum \left( \frac{3}{3} \right)^n=\sum 1$, diverges. So open $(-1,5)$. Yes A. Tempting $[-1,5]$ fails as terms don't go to 0. Verify term limit to 0 at endpoints.