Answer: $\sum_{n=0}^{\infty} (n+1) x^n$

Differentiate the geometric series: Since $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$, differentiating both sides: $\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1}$. Re-index the series: Substituting $m = n - 1$: $\sum_{n=1}^{\infty} n x^{n-1} = \sum_{m=0}^{\infty} (m+1) x^m = \sum_{n=0}^{\infty} (n+1) x^n$. Why distractors fail: Option A represents $\frac{1}{1-x^2}$, not $\frac{1}{(1-x)^2}$. Option C starts at $n=0$ giving a zero first term and is equivalent to Option B only when starting at $n=1$, so as written it is nonstandard. Option D is the integral of the geometric series, representing $-\ln(1-x)$ (up to a constant).

Answer: Differentiate the geometric series and multiply by $x$

Derive $\frac{1}{(1-x)^2}$ first: Differentiating $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ gives $\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1}$. Multiply by $x$: $\frac{x}{(1-x)^2} = x \cdot \sum_{n=1}^{\infty} n x^{n-1} = \sum_{n=1}^{\infty} n x^n$. Why distractors fail: Option A gives $x^2 \cdot \frac{1}{1-x}$, not the desired function. Option C produces antiderivatives, not derivatives, yielding $-\ln(1-x)$ type expressions. Option D gives $\frac{1}{1-x^2}$.

Answer: Obtain a power series for the derivative of the function represented by the original series

Term-by-term differentiation principle: If $f(x) = \sum_{n=0}^{\infty} a_n(x-c)^n$ for $|x-c| < R$, then $f'(x) = \sum_{n=1}^{\infty} n a_n(x-c)^{n-1}$ for the same $|x-c| < R$. The radius of convergence remains the same. Why distractors fail: Option A is wrong because the radius stays the same (though endpoint behavior may change). Option C confuses differentiation with recentering. Option D is incorrect because differentiation does not determine endpoint convergence.

Answer: $[-1, 1)$

Test the left endpoint $x = -1$: At $x = -1$: $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$, which is the alternating harmonic series. This converges by the Alternating Series Test. Test the right endpoint $x = 1$: At $x = 1$: $\sum_{n=1}^{\infty} \frac{1}{n}$, the harmonic series, which diverges. State the interval: Including $x = -1$ and excluding $x = 1$, the interval of convergence is $[-1, 1)$. Why distractors fail: Option A excludes $x = -1$ where the series converges. Option C includes $x = 1$ where the series diverges. Option D includes both endpoints but the harmonic series diverges at $x = 1$.

Answer: $(-2, 2)$

Rewrite as geometric series: $\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n} = \sum_{n=0}^{\infty} \left(\frac{-x}{2}\right)^n$. This is geometric with ratio $r = -x/2$. Find convergence condition: Convergence requires $|r| = \frac{|x|}{2} < 1$, so $|x| < 2$ and $R = 2$. Test endpoints: At $x = 2$: $\sum (-1)^n$, which diverges. At $x = -2$: $\sum 1^n$, which also diverges. So both endpoints are excluded. Why distractors fail: Option B includes both endpoints, but the series diverges there. Option C uses $R = 1$ instead of $R = 2$. Option D incorrectly includes the left endpoint.

Answer: $R = \frac{1}{3}$

Apply the Ratio Test: $\left|\frac{a_{n+1}}{a_n}\right| = 3 \cdot \frac{(n+1)^2}{(n+2)^2} \cdot |x+1| \to 3|x+1|$ as $n \to \infty$. Solve for convergence: $3|x+1| < 1 \implies |x+1| < \frac{1}{3}$, so $R = \frac{1}{3}$. Why distractors fail: Option A confuses $R$ with $1/R$. Option C ignores the factor of $3^n$. Option D squares $3$ without justification.

Answer: The series converges for all real numbers $x$

Interpretation of infinite radius: $R = \infty$ means $|x - c| < R$ is satisfied for every real number $x$. The series converges absolutely for all $x \in (-\infty, \infty)$. Why distractors fail: Option A describes $R = 0$. Option C is incorrect because $R = \infty$ guarantees absolute (not merely conditional) convergence everywhere. Option D describes divergence, which contradicts $R = \infty$.

Answer: $R = \infty$

Apply the Ratio Test: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|2x|^{n+1}}{(n+1)!} \cdot \frac{n!}{|2x|^n} = \frac{2|x|}{n+1}$. Take the limit: As $n \to \infty$, $\frac{2|x|}{n+1} \to 0 < 1$ for all $x$. The series converges for every real number $x$, so $R = \infty$. Context: This series equals $e^{2x}$, which is defined for all real $x$, consistent with $R = \infty$. Why distractors fail: Options A, B, and C all give finite radii. The factorial in the denominator ensures the ratio goes to $0$ for any fixed $x$, giving infinite radius.

Answer: $\sum_{n=0}^{\infty} (-1)^n x^{2n}$

Use the geometric series template: Start with $\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n$. Substitute $u = -x^2$ to get $\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n}$. Convergence condition: This is valid for $|-x^2| < 1$, i.e., $|x| < 1$. Why distractors fail: Option A omits the alternating sign $(-1)^n$. Option C uses $x^n$ instead of $x^{2n}$, which represents $\frac{1}{1+x}$. Option D starts at $n=1$ and has incorrect sign pattern.

Answer: The series converges conditionally at $x = 1$

Substitute the endpoint: At $x = 1$: $(x+3) = 4$, so the series becomes $\sum_{n=1}^{\infty} \frac{(-1)^n 4^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$. Determine convergence type: This is the alternating harmonic series, which converges by the Alternating Series Test. However, the absolute value series $\sum \frac{1}{n}$ diverges (harmonic series), so convergence is conditional. Why distractors fail: Option A is wrong because $\sum 1/n$ diverges so convergence is not absolute. Option B is wrong because the alternating harmonic series converges. Option D is wrong because the Ratio Test gives $L = 1$ at endpoints and is inconclusive.

Answer: $R = 0$

Apply the Ratio Test: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{2^{n+1}} \cdot \frac{2^n}{n!} \cdot |x| = \frac{n+1}{2} |x|$. Take the limit: As $n \to \infty$, $\frac{n+1}{2}|x| \to \infty$ for any $x \neq 0$. The series diverges for all $x \neq 0$, so $R = 0$. Why distractors fail: Options B, C, and D all suggest the series converges on some interval around $0$, but the factorial growth in the numerator overwhelms $2^n$, forcing divergence everywhere except $x = 0$.

Answer: $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$

Standard geometric series formula: The geometric series identity states $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| < 1$. The sum starts at $n = 0$. Why distractors fail: Option A starts at $n = 1$, which omits the constant term $1$ and gives $\frac{x}{1-x}$. Option C equals $\sum_{n=0}^{\infty} (-1)^n x^n$, not $\sum x^n$. Option D represents $\frac{1}{1+x}$, not $\frac{1}{1-x}$.

Answer: $\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n \cdot 5^n}$

Verify the radius for all options: All options with $5^n$ in the denominator have $R = 5$ by the Ratio Test, since the limit involves $\frac{|x|}{5}$ which must be less than $1$. Check Option D at $x = 5$: At $x = 5$: $\sum_{n=1}^{\infty} \frac{(-1)^n 5^n}{n \cdot 5^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$. This is the alternating harmonic series, which converges by the Alternating Series Test. Check Option D at $x = -5$: At $x = -5$: $\sum_{n=1}^{\infty} \frac{(-1)^n (-5)^n}{n \cdot 5^n} = \sum_{n=1}^{\infty} \frac{(-1)^n (-1)^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n}$. This is the harmonic series, which diverges. Why distractors fail: Option A ($\sum x^n/(n \cdot 5^n)$) converges at $x = -5$ (alternating harmonic) and diverges at $x = 5$ (harmonic)—the opposite of what is required. Option B ($\sum x^n/(n^2 \cdot 5^n)$) converges at both endpoints since $\sum 1/n^2$ is a convergent $p$-series. Option C ($\sum x^n/5^n$) is a geometric series that diverges at both endpoints.

Answer: The Root Test, because $\limsup_{n\to\infty} |a_n|^{1/n}$ handles alternating coefficient formulas

Why the Ratio Test struggles here: The ratio $|a_{n+1}/a_n|$ alternates between $3/2$ (when $n$ is even) and $2/3$ (when $n$ is odd), so the limit does not exist. Root Test handles this: $|a_n|^{1/n} = 2$ when $n$ is even and $3$ when $n$ is odd. So $\limsup |a_n|^{1/n} = 3$, giving $R = 1/3$. Why distractors fail: Option A fails because the ratio limit does not exist. Option C is used for series of constants, not to find the radius of a power series. Option D only gives an upper bound, not the exact radius.

Answer: $R = \frac{1}{L}$

Apply the Ratio Test to the general term: The Ratio Test gives $\lim_{n\to\infty} \left|\frac{a_{n+1} x^{n+1}}{a_n x^n}\right| = L|x|$. Convergence requires $L|x| < 1$, so $|x| < \frac{1}{L}$. Identify the radius: The radius of convergence is $R = \frac{1}{L}$ (assuming $0 < L < \infty$). Why distractors fail: Option A confuses $L$ with $R$; they are reciprocals. Options C and D have no basis in the Ratio Test formula.

Answer: $R \geq 3$

Use convergence at $x = 4$: Since the series converges at $x = 4$, we know $|4 - 1| = 3 \leq R$, so $R \geq 3$. Use divergence at $x = -3$: $|-3 - 1| = 4 > R$ would need to hold for divergence there, so $R \leq 4$. Combined: $3 \leq R \leq 4$. Why $R \geq 3$ is the best answer: With only the information given, we can guarantee $R \geq 3$. We cannot pin down the exact value without more information. Why distractors fail: Option A states $R = 3$ exactly, but $R$ could be larger (e.g., $R = 3.5$). Option C states $R = 4$, but divergence at $x = -3$ (distance $4$) shows $R \leq 4$, and $R$ could be less than $4$. Option D contradicts convergence at distance $3$.

Answer: Test convergence separately at $x = c - R$ and $x = c + R$

Ratio Test is inconclusive at endpoints: The Ratio Test yields $L = 1$ at $x = c \pm R$, making it inconclusive at these points. Endpoint testing is necessary: You must substitute $x = c - R$ and $x = c + R$ into the series individually and use other convergence tests (e.g., Alternating Series Test, p-series test) to determine convergence or divergence. Why distractors fail: Option A is wrong because the Ratio Test gives $L=1$ at endpoints and thus cannot decide convergence there. Option C is unnecessary since divergence outside the radius is guaranteed by theory. Option D is only a necessary condition for convergence, not sufficient to determine endpoint behavior.

Answer: No, because $|8 - 2| = 6 > 5 = R$, so $x = 8$ is outside the interval of convergence

Identify as geometric series: This is $\sum_{n=0}^{\infty} \left(\frac{x-2}{5}\right)^n$, geometric with ratio $r = \frac{x-2}{5}$. Find convergence condition: Convergence requires $\left|\frac{x-2}{5}\right| < 1$, i.e., $|x - 2| < 5$, so $R = 5$ and the interval is $(-3, 7)$. Evaluate the claim: At $x = 8$: $|8 - 2| = 6 > 5$, so $x = 8$ is outside the interval and the series diverges there. Why distractors fail: Option A incorrectly states $8$ is within the radius. Option C falsely claims geometric series converge everywhere. Option D understates the radius, which is $5$, not $0$.

Answer: $R = 1$

Recognize the geometric series: The series $\sum_{n=0}^{\infty} x^n$ is a geometric series with common ratio $x$. A geometric series converges when $|x| < 1$. Determine the radius: Since the series converges for $|x| < 1$, the radius of convergence is $R = 1$. Why distractors fail: $R = 0$ would mean the series converges only at the center, which is false. $R = 2$ and $R = \infty$ overstate the interval of convergence.

Answer: $[1, 5]$

Find the radius: Using the Ratio Test: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x-3|}{2} \cdot \frac{n^2}{(n+1)^2} \to \frac{|x-3|}{2}$. Convergence when $\frac{|x-3|}{2} < 1$, giving $R = 2$ and open interval $(1, 5)$. Test $x = 1$ (left endpoint): At $x = 1$: $\sum_{n=1}^{\infty} \frac{(-2)^n}{n^2 \cdot 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$. This converges absolutely because $\sum \frac{1}{n^2}$ is a convergent $p$-series ($p = 2 > 1$). Test $x = 5$ (right endpoint): At $x = 5$: $\sum_{n=1}^{\infty} \frac{2^n}{n^2 \cdot 2^n} = \sum_{n=1}^{\infty} \frac{1}{n^2}$. This is the same convergent $p$-series. Why distractors fail: Options A, C, and D exclude one or both endpoints. Since the series converges at both endpoints, the interval is the closed interval $[1, 5]$.

Answer: Because the Ratio Test yields $L = 1$ at endpoints, which is the boundary between convergence and divergence

Ratio Test at the boundary: At $x = c \pm R$, the ratio $\lim_{n\to\infty}\left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right| = |x-c| \cdot \frac{1}{R} = 1$. The Ratio Test is inconclusive when $L = 1$. Why distractors fail: Option A is incorrect; $L = 1$, not $0$, at endpoints. Option C is wrong because the Ratio Test applies to series with any terms (using absolute values). Option D has no basis; the Ratio Test does not require factorials.

Answer: $|x - c| < R$

Absolute convergence within the radius: A power series converges absolutely for all $x$ satisfying $|x - c| < R$. At the endpoints $|x - c| = R$, the series may converge conditionally, absolutely, or diverge. Why distractors fail: Options A and D incorrectly include the endpoints, where absolute convergence is not guaranteed. Option C states values outside the radius, where the series diverges.

Answer: Series A has $R = \infty$ and Series B has $R = 0$

Analyze Series A: For $\sum \frac{x^n}{n!}$, the Ratio Test gives $\frac{|x|}{n+1} \to 0$ for all $x$. Therefore $R = \infty$. Analyze Series B: For $\sum n! \, x^n$, the Ratio Test gives $(n+1)|x| \to \infty$ for any $x \neq 0$. Therefore $R = 0$. Why distractors fail: Option A ignores the dramatic difference factorials make in the numerator vs. denominator. Options C and D swap or equalize the results incorrectly.

Answer: The radius of convergence remains unchanged, but endpoint behavior may change

Key theorem on differentiation of power series: A fundamental result states that a power series and its term-by-term derivative have the same radius of convergence $R$. Endpoint caveat: Although $R$ is preserved, convergence at the endpoints $x = c \pm R$ may change. For example, a series that converges at an endpoint might produce a derivative series that diverges there. Why distractors fail: Options A, B, and D all incorrectly state that the radius changes. The radius is invariant under term-by-term differentiation and integration.

Answer: $\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$

Start from $\frac{1}{1+x}$: Replace $x$ with $-x$ in the geometric series: $\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n$ for $|x| < 1$. Integrate term by term: $\int \frac{1}{1+x}\,dx = \ln(1+x) + C$. Integrating the series: $\sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} + C$. Setting $x = 0$ gives $C = 0$. Re-index: Shifting the index: $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$. Why distractors fail: Option A starts at $n=0$, which gives a $\frac{1}{0}$ term. Option C omits the alternating sign. Option D is related to $e^x - 1$, not $\ln(1+x)$.

Answer: $R = 3$

Apply the Ratio Test: Let $a_n = \frac{x^n}{n \cdot 3^n}$. Then $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|}{3} \cdot \frac{n}{n+1}$. As $n \to \infty$, this approaches $\frac{|x|}{3}$. Solve for convergence: For convergence, $\frac{|x|}{3} < 1$, so $|x| < 3$. Therefore $R = 3$. Why distractors fail: $R = 1$ ignores the factor $3^n$ in the denominator. $R = \frac{1}{3}$ confuses $R$ with $\frac{1}{R}$. $R = 9$ has no justification from the Ratio Test.

Answer: The radius is correct, but the student failed to test endpoints, so the interval may include one or both endpoints

Ratio Test was applied correctly: The Ratio Test correctly gives $\lim |x| \cdot \frac{n}{n+1} = |x|$, yielding $R = 1$. Missing endpoint analysis: At $x = 1$: $\sum 1/n$ diverges. At $x = -1$: $\sum (-1)^n/n$ converges. The actual interval is $[-1, 1)$, not $(-1, 1)$. Why distractors fail: Option A is wrong because the left endpoint was missed. Option C is wrong because the Ratio Test application was valid. Option D is wrong because the Ratio Test never determines endpoint convergence.