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

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Question

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

Tap to reveal answer

Answer

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:

$(-\infty, \infty)$

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Question

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

Tap to reveal answer

Answer

Using the root test

and

. T

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

This occurs when x=5.

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Question

Which of following intervals of convergence cannot exist?

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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:

$(-\infty, \infty)$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

Using the root test

and

. T

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

This occurs when x=5.

← Didn't Know|Knew It →

Question

Which of following intervals of convergence cannot exist?

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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:

$(-\infty, \infty)$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

Using the root test

and

. T

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

This occurs when x=5.

← Didn't Know|Knew It →

Question

Which of following intervals of convergence cannot exist?

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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:

$(-\infty, \infty)$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

Using the root test

and

. T

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

This occurs when x=5.

← Didn't Know|Knew It →

Question

Which of following intervals of convergence cannot exist?

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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:

$(-\infty, \infty)$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

Using the root test

and

. T

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

This occurs when x=5.

← Didn't Know|Knew It →

Question

Which of following intervals of convergence cannot exist?

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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:

$(-\infty, \infty)$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

Using the root test

and

. T

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

This occurs when x=5.

← Didn't Know|Knew It →

Question

Which of following intervals of convergence cannot exist?

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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:

$(-\infty, \infty)$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

Using the root test

and

. T

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

This occurs when x=5.

← Didn't Know|Knew It →

Knew It

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 }$$$\frac{3^nx$^n}{n!}$$ .

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:

$(-\infty, \infty)$

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

Using the root test

and

. T

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

This occurs when x=5.

$\small [2,\infty)$ 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 }$$$\frac{3^nx$^n}{n!}$$ .

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:

$(-\infty, \infty)$

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

Using the root test

and

. T

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

This occurs when x=5.

$\small [2,\infty)$ 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 }$$$\frac{3^nx$^n}{n!}$$ .

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:

$(-\infty, \infty)$

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

Using the root test

and

. T

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

This occurs when x=5.

$\small [2,\infty)$ 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 }$$$\frac{3^nx$^n}{n!}$$ .

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:

$(-\infty, \infty)$

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

Using the root test

and

. T

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

This occurs when x=5.

$\small [2,\infty)$ 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 }$$$\frac{3^nx$^n}{n!}$$ .

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:

$(-\infty, \infty)$

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

Using the root test

and

. T

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

This occurs when x=5.

$\small [2,\infty)$ 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 }$$$\frac{3^nx$^n}{n!}$$ .

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:

$(-\infty, \infty)$

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

Using the root test

and

. T

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

This occurs when x=5.

$\small [2,\infty)$ 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 }$$$\frac{3^nx$^n}{n!}$$ .

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:

$(-\infty, \infty)$

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

Using the root test

and

. T

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

This occurs when x=5.