Radius and Interval of Convergence of Power Series - AP Calculus BC

Card 0 of 30

Question

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Answer

Using the root test,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0$

Because 0 is always less than 1, the root test shows that the series converges for any value of x.

Therefore, the interval of convergence is:

$(-\infty, \infty)$

Compare your answer with the correct one above

Question

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

Answer

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

Compare your answer with the correct one above

Question

Which of following intervals of convergence cannot exist?

Answer

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Compare your answer with the correct one above

Question

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Answer

Using the root test,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0$

Because 0 is always less than 1, the root test shows that the series converges for any value of x.

Therefore, the interval of convergence is:

$(-\infty, \infty)$

Compare your answer with the correct one above

Question

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

Answer

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

Compare your answer with the correct one above

Question

Which of following intervals of convergence cannot exist?

Answer

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Compare your answer with the correct one above

Question

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Answer

Using the root test,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0$

Because 0 is always less than 1, the root test shows that the series converges for any value of x.

Therefore, the interval of convergence is:

$(-\infty, \infty)$

Compare your answer with the correct one above

Question

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

Answer

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

Compare your answer with the correct one above

Question

Which of following intervals of convergence cannot exist?

Answer

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Compare your answer with the correct one above

Question

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Answer

Using the root test,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0$

Because 0 is always less than 1, the root test shows that the series converges for any value of x.

Therefore, the interval of convergence is:

$(-\infty, \infty)$

Compare your answer with the correct one above

Question

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

Answer

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

Compare your answer with the correct one above

Question

Which of following intervals of convergence cannot exist?

Answer

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Compare your answer with the correct one above

Question

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Answer

Using the root test,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0$

Because 0 is always less than 1, the root test shows that the series converges for any value of x.

Therefore, the interval of convergence is:

$(-\infty, \infty)$

Compare your answer with the correct one above

Question

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

Answer

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

Compare your answer with the correct one above

Question

Which of following intervals of convergence cannot exist?

Answer

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Compare your answer with the correct one above

Question

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Answer

Using the root test,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0$

Because 0 is always less than 1, the root test shows that the series converges for any value of x.

Therefore, the interval of convergence is:

$(-\infty, \infty)$

Compare your answer with the correct one above

Question

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

Answer

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

Compare your answer with the correct one above

Question

Which of following intervals of convergence cannot exist?

Answer

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Compare your answer with the correct one above

Question

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Answer

Using the root test,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0$

Because 0 is always less than 1, the root test shows that the series converges for any value of x.

Therefore, the interval of convergence is:

$(-\infty, \infty)$

Compare your answer with the correct one above

Question

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

Answer

Using the root test

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$

and

$\lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$ . T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

Compare your answer with the correct one above

Tap the card to reveal the answer

Radius and Interval of Convergence of Power Series - AP Calculus BC

Find the interval of convergence of for the series .

Using the root test, Because 0 is always less than 1, the root test shows that the series converges for any value of x. Therefore, the interval of convergence is:

Find the interval of convergence for of the Taylor Series .

Using the root test and . T herefore, the series only converges when it is equal to zero. This occurs when x=5.

Which of following intervals of convergence cannot exist?

cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence ). Thus, can never be an interval of convergence.

Find the interval of convergence of for the series .

Using the root test, Because 0 is always less than 1, the root test shows that the series converges for any value of x. Therefore, the interval of convergence is:

Find the interval of convergence for of the Taylor Series .

Using the root test and . T herefore, the series only converges when it is equal to zero. This occurs when x=5.

Which of following intervals of convergence cannot exist?

cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence ). Thus, can never be an interval of convergence.

Find the interval of convergence of for the series .

Using the root test, Because 0 is always less than 1, the root test shows that the series converges for any value of x. Therefore, the interval of convergence is:

Find the interval of convergence for of the Taylor Series .

Using the root test and . T herefore, the series only converges when it is equal to zero. This occurs when x=5.

Which of following intervals of convergence cannot exist?

cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence ). Thus, can never be an interval of convergence.

Find the interval of convergence of for the series .

Using the root test, Because 0 is always less than 1, the root test shows that the series converges for any value of x. Therefore, the interval of convergence is:

Find the interval of convergence for of the Taylor Series .

Using the root test and . T herefore, the series only converges when it is equal to zero. This occurs when x=5.

Which of following intervals of convergence cannot exist?

cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence ). Thus, can never be an interval of convergence.

Find the interval of convergence of for the series .

Using the root test, Because 0 is always less than 1, the root test shows that the series converges for any value of x. Therefore, the interval of convergence is:

Find the interval of convergence for of the Taylor Series .

Using the root test and . T herefore, the series only converges when it is equal to zero. This occurs when x=5.

Which of following intervals of convergence cannot exist?

cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence ). Thus, can never be an interval of convergence.

Find the interval of convergence of for the series .

Using the root test, Because 0 is always less than 1, the root test shows that the series converges for any value of x. Therefore, the interval of convergence is:

Find the interval of convergence for of the Taylor Series .

Using the root test and . T herefore, the series only converges when it is equal to zero. This occurs when x=5.

Which of following intervals of convergence cannot exist?

cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence ). Thus, can never be an interval of convergence.

Find the interval of convergence of for the series .

Using the root test, Because 0 is always less than 1, the root test shows that the series converges for any value of x. Therefore, the interval of convergence is:

Find the interval of convergence for of the Taylor Series .

Using the root test and . T herefore, the series only converges when it is equal to zero. This occurs when x=5.

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

Find the interval of convergence of for the series $\sum_{n=1}^{\infty }\frac{3^nx^n}{n!}$ .

Find the interval of convergence for of the Taylor Series $\sum_{n=0}^{\infty }n!(x-5)^n$ .