Question 1

Find the interval of convergence and radius of convergence of the power series: $$\sum _ { k = 0 } ^ { \infty } \frac { 1 } { 1 + 3 ^ { k } } x ^ { k }$$&#10;A) R = 1, interval of convergence: (-1, 1)&#10;B) R = 1, interval of convergence: (-3, 3)&#10;C) R = 3, interval of convergence: (-1, 1)&#10;D) R = 3, interval of convergence: (-3, 3)

Accepted Answer

R = 3, interval of convergence: (-3, 3)&#10;

Question 2

Find the interval of convergence and radius of convergence of the power series: $$\sum _ { k = 0 } ^ { \infty } \frac { 1 } { 1 + 5 ^ { k } } x ^ { k }$$&#10;A) R = 1, interval of convergence: (3, 5)&#10;B) R = 3, interval of convergence: (-3, -1)&#10;C) R = 5, interval of convergence: (-5, 5)&#10;D) R = 5, interval of convergence: (-5, -3)

Accepted Answer

The radius of convergence (R) for a power series can be found using the formula $R = \lim_{k \to \infty} \left| \frac{a_k}{a_{k+1}} \right|$, where $a_k$ is the coefficient of $x^k$. For the given series, $a_k = \frac{1}{1+5^k}$. The ratio of successive terms is $\frac{1+5^{k+1}}{1+5^k} = 5$, indicating that the radius of convergence is $R = \frac{1}{5}$. However, this method was incorrectly applied here. The correct approach for finding the radius of convergence for this series involves recognizing that it resembles a geometric series with a common ratio related to $x$. The series converges when $|x| < 5$, hence the radius of convergence is $R = 5$, and the interval of convergence is $(-5, 5)$, considering the series is a geometric series with a ratio less than 1 in absolute value for $|x| < 5$.

Question 3

Find the interval of convergence and radius of convergence of the power series: $$\sum _ { k = 0 } ^ { \infty } ( k + 1 ) ! x ^ { k }$$&#10;A) R = 1, interval of convergence: (-1, 1)&#10;B) R = 0, series converges only when x = 1&#10;C) R = 0, series converges only when x = 0&#10;D) R = 1, interval of convergence: (0, 2)

Accepted Answer

The radius of convergence (R) can be found using the ratio test, which involves taking the limit as $k$ approaches infinity of the absolute value of the ratio of the $(k+1)$th term to the $k$th term of the series. For the given series, $\sum _ { k = 0 } ^ { \infty } ( k + 1 ) ! x ^ { k }$, this ratio is $\lim_{k \to \infty} \frac{(k+2)!|x|^{k+1}}{(k+1)!|x|^k} = \lim_{k \to \infty} (k+2)|x|$, which approaches infinity for any $x$ other than 0. This means the series only converges when $x = 0$, making the radius of convergence $R = 0$.

Question 4

Find the interval of convergence and radius of convergence of the power series: $$\sum _ { k = 0 } ^ { \infty } \left( \frac { 3 } { k } \right) ^ { k } x ^ { 2 k }$$&#10;A) R = 1, interval of convergence: (-1, 1)&#10;B) R = ?, interval of convergence: (-?, ?)&#10;C) R = 1, interval of convergence: (0, 2)&#10;D) R = 3, interval of convergence: (-3, 3)

Accepted Answer

The answer of Find the interval of convergence and radius...

Question 5

Find the interval of convergence and radius of convergence of the power series: $$\sum _ { k = 0 } ^ { \infty } \left( \frac { 3 } { k } \right) ^ { k } x ^ { 3 k }$$&#10;A) R = 1, interval of convergence: (-1, 1)&#10;B) R = ?, interval of convergence: (-?, ?)&#10;C) R = 1, interval of convergence: (0, 2)&#10;D) R = 3, interval of convergence: (-3, 3)

Accepted Answer

To find the radius of convergence (R) and the interval of convergence of the given power series, we use the root test (also known as the nth root test). The root test involves taking the limit of the nth root of the absolute value of the nth term of the series as n approaches infinity. For the given series, the general term is $\left( \frac{3}{k} \right)^k x^{3k}$. Applying the root test, we calculate:$$\lim_{k \to \infty} \left| \left( \frac{3}{k} \right)^k x^{3k} \right|^{1/k} = \lim_{k \to \infty} \left| \frac{3}{k} \cdot x^3 \right| = |x^3| \lim_{k \to \infty} \frac{3}{k} = 0 \cdot |x^3| = 0$$This calculation does not directly provide the radius of convergence in the usual manner because the limit does not depend on $x$ in the expected way to give us a comparison to $1$ (as in $|x| < R$). The mistake in the calculation suggests a misunderstanding in applying the root test correctly for this series. The correct approach should involve recognizing that the series' form does not fit the standard criteria for applying the root or ratio tests directly to find the radius of convergence in a straightforward manner. Given the error in the initial explanation and the unique nature of the series, the correct answer should involve a more detailed analysis specific to the series' form. However, without the correct application of convergence tests and considering the options provided, none of the given intervals or radius of convergence values directly follow from the incorrect application of the root test as described. The correct method involves recognizing the series diverges for all $x$ except possibly at $x = 0$, due to the terms not tending to zero as $k$ increases for any $x \neq 0$. Thus, the initial explanation does not accurately lead to a correct choice among the provided options, and a more thorough analysis is required to determine the series' behavior accurately.

Question 6

Find the interval of convergence of the power series: $$\sum _ { k = 0 } ^ { \infty } \frac { 1 } { 2 k ! } x ^ { k }$$&#10;A) (-1, 1)&#10;B) Converges only when x = 0.&#10;C) (0, 2)&#10;D) (-?, ?)

Accepted Answer

The series is a power series of the form $\sum _ { k = 0 } ^ { \infty } a_k x ^ { k }$ where $a_k = \frac{1}{2k!}$. The radius of convergence R can be found using the ratio test, which in this case, due to the factorial in the denominator, leads to R being infinite. This means the series converges for all real numbers, hence the interval of convergence is $(-∞, ∞)$.

Question 7

Find the interval of convergence of the power series: $$\sum _ { k = 0 } ^ { \infty } \frac { x ^ { k } } { ( 2 k ) ! }$$&#10;A) (-1, 1)&#10;B) (-?, ?)&#10;C) (0, 2)&#10;D) Converges only when x = 0

Accepted Answer

The series converges for all x, which is indicated by option B with the symbol "?", meaning there are no bounds on the interval of convergence. This is determined by the Ratio Test, where the limit of the absolute value of the ratio of consecutive terms as k approaches infinity is 0 for any finite x, due to the factorial in the denominator growing much faster than any power of x.

Question 8

Find the interval of convergence of the power series: $\sum _ { k = 0 } ^ { \infty } \frac { ( - 3 ) ^ { k } } { k ! } x ^ { k }$&#10;A) (-1, 1)&#10;B) (0, 2)&#10;C) (-?, ?)&#10;D) Converges only when x = 0

Accepted Answer

The series is a Taylor series expansion of $e^{-3x}$, which converges for all real numbers x, hence the interval of convergence is $(-∞, ∞)$.

Question 9

Find the interval of convergence of the power series: $$\sum _ { k = 0 } ^ { \infty } \frac { ( - 1 ) ^ { k } } { ( 3 k ) ! } x ^ { 2 k }$$&#10;A) (-1, 1)&#10;B) (0, 2)&#10;C) (-?, ?)&#10;D) ( 0, 1)

Accepted Answer

The series is an alternating series with factorial in the denominator, which suggests rapid growth of the denominator as $k$ increases. The factorial growth in the denominator typically leads to a power series that converges for all $x$, hence the interval of convergence is $(-∞, ∞)$.

Question 10

Find the interval of convergence of the power series: $\sum _ { k = 0 } ^ { \infty } \frac { ( - 1 ) ^ { k } } { k + 2 } x ^ { k }$&#10;A) (-1, 1)&#10;B) (0, 1)&#10;C) [-1, 1]&#10;D) (-1, 1]

Accepted Answer