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 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 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.

The skill here is finding the radius and interval of convergence for a power series. To find the radius, apply the ratio test: the limit as n approaches infinity of $ \left| \frac{3^{n+1} (x-1)^{n+1}}{(n+1) 4^{n+1}} \right| / \left| \frac{3^n (x-1)^n}{n 4^n} \right| = |x-1| \cdot \frac{3}{4} \cdot \frac{n}{n+1} \to |x-1| \cdot \frac{3}{4} < 1 $, so $ |x-1| < \frac{4}{3} $, giving $ R = \frac{4}{3} $. Since the question asks only for the radius, we stop here. The weighted ratio $ \frac{3}{4} $ inverts to $ R = \frac{4}{3} $. A tempting distractor is $ R = \frac{3}{4} $, but this fails because R is the reciprocal of the limit coefficient. 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. Center at -1, ratio $|x+1|$ lim $\frac{1}{n+1}$ wait no, it's $\sum \frac{(x+1)^n}{n^2}$, ratio $| \frac{a_{n+1}}{a_n} | = |x+1| \cdot \frac{n^2}{(n+1)^2} \to |x+1| < 1$? Wait, lim $= |x+1|$, so converges when $|x+1| < 1$, R=1, open $(-2,0)$. But check endpoints: at x=-2, $\sum \frac{(-1)^n}{n^2}$, alternates, converges absolutely actually since $\sum \frac{1}{n^2}$ converges. At x=0, $\sum \frac{1^n}{n^2} = \sum \frac{1}{n^2}$ converges. So $[-2,0]$. Yes. Tempting $(-2,0)$ fails to recognize p-series convergence at endpoints. Use absolute convergence tests at boundaries for inclusion.

The skill being tested is finding the radius of convergence for a power series. Ratio test: lim $|n^2$ $(x-1)^n$ / $5^n$ * $5^{n+1}$ / $((n+1)^2$ $(x-1)^{n+1}$) | wait, actually lim $|(a_{n+1}$/a_n)| = |x-1|/5 * lim $(n+1)^2$ / $n^2$ = |x-1|/5 <1, so |x-1|<5, R=5. The $n^2$ grows polynomially, overpowered by exponential $5^n$. No endpoints here as question is radius only. Tempting R=1/5 might come from inverting the limit incorrectly. Remember to simplify the limit properly in ratio test for accurate radius in various series.

The skill here is finding the radius and interval of convergence for a power series. To find the radius, apply the ratio test: the limit as n approaches infinity of $|(x-3)^{n+1}/(n 2^{n+1})| / |(x-3)^n/((n-1) 2^n)| = $|x-3|/2 \cdot ((n-1)/n) \rightarrow |x-3|/2 < 1$, so $|x-3| < 2$, giving $R=2$. Since the question asks only for the radius, we stop here, but endpoints would need checking for the full interval. The shift in indexing from n=2 does not affect the limit. A tempting distractor is $R=1/2$, but this fails because the 2^n in the denominator gives $R=2$. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.

Finding the radius of convergence requires applying the ratio test to this power series. We compute $\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}| = \lim_{n\to\infty}|\frac{(x-5)n}{4(n+1)}| = \frac{|x-5|}{4}\lim_{n\to\infty}\frac{n}{n+1} = \frac{|x-5|}{4}$. The series converges when $\frac{|x-5|}{4} < 1$, which means $|x-5| < 4$, giving radius $R = 4$. The interval before checking endpoints would be $(1, 9)$. A common mistake is to think the denominator $4^n$ means $R = \frac{1}{4}$, but the ratio test shows the radius equals the reciprocal of the coefficient in the limit. Remember that for series of the form $\sum \frac{(x-c)^n}{a^n}$, the radius of convergence is $R = a$.