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

Card 0 of 30

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

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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 }$$n!(x-5)^n$$ .

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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?

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

Tap to see back →

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 }$$n!(x-5)^n$$ .

Tap to see back →

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?

Tap to see back →

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

Tap to see back →

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 }$$n!(x-5)^n$$ .

Tap to see back →

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?

Tap to see back →

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

Tap to see back →

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 }$$n!(x-5)^n$$ .

Tap to see back →

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?

Tap to see back →

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

Tap to see back →

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 }$$n!(x-5)^n$$ .

Tap to see back →

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?

Tap to see back →

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

Tap to see back →

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 }$$n!(x-5)^n$$ .

Tap to see back →

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?

Tap to see back →

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

Tap to see back →

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 }$$n!(x-5)^n$$ .

Tap to see back →

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.