Power Series: Radius, Interval of Convergence - AP Calculus BC

Find the radius of convergence of $\sum_{n=0}^{\infty} \dfrac{(x+2)^n}{5^n\sqrt{n+1}}$ .
Answer: $R=5$ Finding the interval of convergence requires determining the radius of convergence for this power series. Using the ratio test, we compute $\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}| = \lim_{n\to\infty}|\frac{(x+2)\sqrt{n+1}}{5\sqrt{n+2}}| = \frac{|x+2|}{5}$ . The series converges when $\frac{|x+2|}{5} < 1$ , which means $|x+2| < 5$ , giving radius $R = 5$ . The interval before checking endpoints is $(-7, 3)$ . A common mistake is to think the $\frac{1}{5^n}$ factor means $R = \frac{1}{5}$ , but the ratio test shows the radius is actually the reciprocal of the coefficient of $(x+2)$ in the limit. Remember that the radius of convergence is the value $R$ such that the series converges for $|x-c| < R$ where $c$ is the center.
Compute the interval of convergence of $\sum_{n=2}^{\infty}\frac{(x-4)^n}{n(n-1)}$ for an error-correction series.
Answer: [3,5] The skill being tested is finding the interval of convergence for a power series. Center at 4, ratio test on $(x-4)^n / (n(n-1))$ gives $|x-4| \lim \frac{n(n-1)}{(n+1)n} = |x-4| < 1$ ? Wait, $\lim \frac{n(n-1)}{(n+1)n} = 1$ , so $|x-4| < 1$ , radius 1, open $(3,5)$ . But check endpoints: at x=3, $\sum (-1)^n / (n(n-1))$ , partial fractions $1/(n(n-1)) = 1/(n-1) - 1/n$ , so $\sum (-1)^n (1/(n-1) - 1/n)$ , which is alternating telescoping, converges. At x=5, $(1)^n / n(n-1)$ , positive, like $1/n^2$ , converges by p-series. Wait, yes both converge. So $[3,5]$ . Yes marked B. Tempting $(3,5)$ fails to check endpoint convergence via partial fractions. Use decomposition for rational terms to assess endpoint behavior in strategy.
What is the radius of convergence of $\sum_{n=1}^{\infty}\frac{(x+4)^n}{n\,6^n}$ in a computation of $\ln$ -like behavior?
Answer: $R=6$ The skill being tested is finding the radius of convergence for a power series. Ratio $\frac{|x+4|}{6}$ lim $\frac{1}{n+1} \cdot n$ ? Wait, term $\frac{(x+4)^n}{n \, 6^n}$ , ratio $\frac{|x+4|}{6} \cdot \frac{n}{n+1} \to \frac{|x+4|}{6} <1$ , R=6. Like log series. No endpoints. Tempting R=1/6 inverts incorrectly. Simplify limit before concluding.
What is the radius of convergence of $\sum_{n=0}^{\infty}\frac{(x+1)^{n}}{7^{n}}$ for a geometric-series approximation?
Answer: $R=7$ 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.
What is the radius of convergence of $\sum_{n=1}^{\infty} \dfrac{n!}{3^n}(x+4)^n$ ?
Answer: $R=0$ Finding the radius of convergence for factorial coefficient series requires careful ratio test application. We compute $\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}| = \lim_{n\to\infty}|\frac{(n+1)!(x+4)3^n}{n!3^{n+1}}| = \lim_{n\to\infty}\frac{(n+1)|x+4|}{3} = \infty$ for any $x \neq -4$ . Since this limit exceeds 1 for all $x \neq -4$ , the series diverges everywhere except possibly at $x = -4$ . At $x = -4$ , all terms are zero, so the series converges trivially to 0, giving radius $R = 0$ . A common mistake is thinking factorial numerators might allow some interval of convergence, but factorials in numerators grow too fast for convergence. The key principle is that factorial growth in numerators overwhelms any exponential decay, limiting convergence to a single point.
Determine the radius of convergence of $\sum_{n=0}^{\infty}\dfrac{(x-2)^n}{n!}$ .
Answer: $R=\infty$ This problem involves finding the radius of convergence for the exponential series. The series $\sum_{n=0}^{\infty}\frac{(x-2)^n}{n!}$ is the Taylor series for $e^{x-2}$ . Using the ratio test, $\lim_{n\to\infty}\left|\frac{(x-2)\cdot n!}{(n+1)!}\right| = \lim_{n\to\infty}\frac{|x-2|}{n+1} = 0$ for any finite $x$ . Since this limit is always 0 < 1, the series converges for all real $x$ , giving radius $R=\infty$ . Students might think $R=1$ by analogy with geometric series, but factorial growth in the denominator overwhelms any polynomial growth in the numerator. The exponential function's Taylor series converges everywhere, illustrating that some power series have infinite radius of convergence.
Find the radius of convergence of $\sum_{n=1}^{\infty} \dfrac{(2x+1)^n}{n\,7^n}$ .
Answer: $R=\tfrac{7}{2}$ Finding the radius of convergence requires careful handling of the linear term in the power series. We can rewrite the series as $\sum_{n=1}^{\infty} \frac{2^n(x+\frac{1}{2})^n}{n\cdot 7^n}$ , and applying the ratio test gives $\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}| = \frac{2|x+\frac{1}{2}|}{7}$ . The series converges when $\frac{2|x+\frac{1}{2}|}{7} < 1$ , which means $|x+\frac{1}{2}| < \frac{7}{2}$ , so the radius is $R = \frac{7}{2}$ . The interval before checking endpoints is $(-4, 3)$ . A common error is to think the coefficient 2 in $(2x+1)^n$ doesn't affect the radius, but it appears in the ratio test calculation. Remember that for $(ax+b)^n$ terms, you must factor out the coefficient $a$ and account for it when finding the radius.
What is the radius of convergence of $\sum_{n=0}^{\infty}\frac{(x-1)^{3n}}{8^n}$ in a compact encoding scheme?
Answer: $R=2$ 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 | $(x-1)^{3(n+1)}$ / $8^{n+1}$ | / | $(x-1)^{3n}$ / $8^n$ | = | $(x-1)^3$ |/8 < 1, so |x-1| < 2, giving R=2. Since the question asks only for the radius, we stop here, but note the interval would be |x-1| < 2 with endpoints to check if needed. The powers are multiples of 3, but the ratio test still applies effectively. A tempting distractor is R=∛8 (which is also 2), but the simplified form is R=2. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.
What is the interval of convergence for $\sum_{n=0}^{\infty}\frac{(x-10)^n}{(-5)^n}$ used in a geometric update rule?
Answer: $(5,15)$ The skill here is finding the interval of convergence for a power series. Recognize this as a geometric series $\sum [-(x-10)/5]^n$ , which converges when $|- (x-10)/5| < 1$ , so $|x-10| < 5$ , giving open interval $(5, 15)$ . Then, check the endpoints: at x=5, it becomes $\sum 1^n$ , which diverges; at x=15, $\sum (-1)^n$ , which oscillates and diverges. Neither endpoint converges. A tempting distractor is $[5,15]$ , 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.
Determine the radius of convergence for $\sum_{n=0}^{\infty}\dfrac{n!}{(3n)!}(x+2)^n$ .
Answer: $R=\infty$ The skill being tested here is finding the radius of convergence for a power series. Apply the ratio test to n!/(3n)! $(x+2)^n$ , resulting in a limit of |x+2| / [ (3n+1)(3n+2)(3n+3) ] times (n+1), which approaches 0 as n goes to infinity. This implies convergence for all x, so the radius is infinity. The factorial in the denominator grows much faster, ensuring this outcome. A tempting distractor like R=1/3 fails by misapplying the ratio without proper limit calculation. Always remember to use the ratio or root test for the radius and separately test endpoints for conditional convergence in power series problems.
For $f(x)=\sum_{n=1}^{\infty}\dfrac{(x-2)^n}{n\,3^n}$ , what is the interval of convergence?
Answer: $( -1,5]$ The skill being tested here is finding the interval of convergence for a power series. To determine the radius of convergence, apply the ratio test to the general term, yielding a limit of |x-2|/3, which is less than 1 when |x-2| < 3, so the radius is 3 and the open interval is (-1,5). Next, check the endpoints: at x=5, the series becomes the harmonic series ∑1/n, which diverges. At x=-1, it becomes the alternating harmonic series ∑ $(-1)^n$ /n, which converges conditionally. A tempting distractor like (-1,5) fails because it excludes x=-1, where the series actually converges due to alternating signs. Always remember to use the ratio or root test for the radius and separately test endpoints for conditional convergence in power series problems.
Determine the radius of convergence of $\sum_{n=1}^{\infty}\frac{(2x-3)^n}{n^3}$ for a numerical series library.
Answer: $R=\frac12$ The skill being tested is finding the radius of convergence for a power series. Term $(2x-3)^n / n^3$ , ratio $|2x-3| \lim \frac{n^3}{(n+1)^3} \to |2x-3| < 1$ , so $|2x-3| < 1$ , $|x - \frac{3}{2}| < \frac{1}{2}$ , $R=\frac{1}{2}$ . The coefficient 2 halves the radius. No endpoints. Tempting $R=2$ ignores the 2 in numerator. Account for linear coefficients in the base for accurate R.
What is the radius of convergence of $\sum_{n=0}^{\infty}\frac{(x-1)^n}{2^n\,n!}$ for a Taylor approximation engine?
Answer: $R=\infty$ 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 $\left| \frac{(x-1)^{n+1}}{2^{n+1} (n+1)!} \right| / \left| \frac{(x-1)^n}{2^n n!} \right| = \frac{|x-1|}{2} \cdot \frac{1}{n+1} \to 0 < 1$ for all x, so $R=\infty$ . The factorial ensures convergence everywhere. This resembles the exponential series scaled. A tempting distractor is $ R=2 $, but this fails because the n! makes the limit 0. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.
Determine the radius of convergence of $\sum_{n=1}^{\infty}\frac{(x+1)^n}{\sqrt{n}\,5^n}$ for an approximation routine.
Answer: $R=5$ 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 $\left| \frac{(x+1)^{n+1}}{\sqrt{n+1} \, 5^{n+1}} \right| / \left| \frac{(x+1)^n}{\sqrt{n} \, 5^n} \right| = \frac{|x+1|}{5} \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \to \frac{|x+1|}{5} < 1$ , so $|x+1| < 5$ , giving R=5. Since the question asks only for the radius, we stop here, but endpoints would be checked for the interval. The square root in the denominator affects the limit negligibly. A tempting distractor is R=√5, but this fails because the $5^n$ dominates, giving R=5. In general, always determine the radius first with the ratio or root test, then separately test each endpoint for conditional convergence.
Determine the radius of convergence of $\sum_{n=1}^{\infty}\frac{3^n(x-1)^n}{n\,4^n}$ for a weighted series model.
Answer: $R=\frac{4}{3}$ 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.
Determine the radius of convergence of $\sum_{n=1}^{\infty} \dfrac{(x-5)^n}{n\,4^n}$ .
Answer: $R=4$ 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 $R = a$ .
Determine the radius of convergence of $\sum_{n=2}^{\infty}\frac{(x-3)^n}{(n-1)2^n}$ for an iterative method.
Answer: $R=2$ 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.
Find the radius of convergence of $\sum_{n=0}^{\infty}\dfrac{(x+2)^n}{5^n\sqrt{n+1}}$ .
Answer: $R=5$ This problem requires finding the radius of convergence for a power series centered at $x=-2$ . Applying the ratio test to the general term $\frac{(x+2)^n}{5^n\sqrt{n+1}}$ , we compute $\lim_{n\to\infty}\left|\frac{(x+2)\sqrt{n+1}}{5\sqrt{n+2}}\right| = \frac{|x+2|}{5}$ . For convergence, this limit must be less than 1, giving $|x+2|<5$ , so the radius is $R=5$ . The factor $\sqrt{n+1}$ in the denominator doesn't affect the radius calculation since $\lim_{n\to\infty}\frac{\sqrt{n+1}}{\sqrt{n+2}}=1$ . Students might incorrectly choose $R=\frac{1}{5}$ by inverting the coefficient, but the ratio test clearly shows we divide by 5, not multiply. Remember that when coefficients have the form $c^n$ , the radius equals $\frac{1}{|c|}$ if in the numerator, or $|c|$ if in the denominator.
Find the radius of convergence of $\sum_{n=0}^{\infty}(-1)^n\frac{(x+5)^{2n}}{9^n}$ .
Answer: $R=3$ This problem involves a power series with even powers only, requiring special attention to the substitution. The series $\sum_{n=0}^{\infty}(-1)^n\frac{(x+5)^{2n}}{9^n}$ can be rewritten as $\sum_{n=0}^{\infty}\frac{(-(x+5)^2)^n}{9^n}$ , which is a geometric series in the variable $u = -(x+5)^2/9$ . This converges when $|u| < 1$ , giving $|(x+5)^2|/9 < 1$ , or $(x+5)^2 < 9$ , which means $|x+5| < 3$ . Therefore, the radius of convergence is $R=3$ . Students often mistakenly think the $2n$ exponent gives radius $\sqrt{3}$ or that the $9^n$ gives radius 9. The key insight is to recognize when a substitution reduces the series to a standard form, then work backwards to find the radius in terms of the original variable.
Determine the interval of convergence for $\sum_{n=0}^{\infty}(-1)^n\dfrac{(x-2)^n}{4^n}$ .
Answer: $(-2,6)$ The skill being tested here is finding the interval of convergence for a power series. This is a geometric series with ratio $r = -(x-2)/4$ , converging when $|r| < 1$ , or $|x-2| < 4$ , giving $(-2,6)$ . At the endpoints $x=-2$ and $x=6$ , the series oscillates and diverges as $|r|=1$ . Geometric series only converge inside the open interval. A tempting distractor like $[-2,6]$ fails because it includes endpoints where the terms don't approach zero. Always remember to use the ratio or root test for the radius and separately test endpoints for conditional convergence in power series problems.

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