Question

Find the radius of convergence, then determine the interval of convergence

Answer 1

The radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series. This can be obtained by using ratio test.  

Find the radius of convergence R and the interval of convergence:

Ratio test is the test that is used to find the convergence of the given power series.  

First aₙ is noted and then aₙ₊₁ is noted.

For  ∑ aₙ,  aₙ and aₙ₊₁ is noted.

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }|$ = β

• If β < 1, then the series converges

• If β > 1, then the series diverges

• If β = 1, then the series inconclusive

Here $a_{k}$ = $\frac{(x+2)^{k}}{\sqrt{k} }$  and  $a_{k+1}$ = $\frac{(x+2)^{k+1}}{\sqrt{k+1} }$

   

Now limit is taken,

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }|$ = $\lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|$

= $\lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|$

= $\lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|$

= $|{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}$

= $|{x+2 }|$ < 1

- 1 < ${x+2 }$ < 1

- 1 - 2 < x < 1 - 2

- 3 < x < - 1

 

We get that,

interval of convergence = (-3, -1)

radius of convergence R = 1

Hence the radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series.

Learn more about radius of convergence here:

brainly.com/question/14394994

#SPJ1

Answer 2

By the ratio test, the series converges for

$\displaystyle \lim_{k\to\infty} \left|\frac{(x+2)^{k+1}}{\sqrt{k+1}} \cdots \frac{\sqrt k}{(x+2)^k}\right| = |x+2| \lim_{k\to\infty} \sqrt{\frac k{k+1}} = |x+2| < 1$

so that the radius of convergence is 1, and the interval of convergence is

|x + 2| < 1   ⇒   -1 < x + 2 < 1   ⇒   -3 < x < -1