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.
[tex]\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }|[/tex] = β
- If β < 1, then the series converges
- If β > 1, then the series diverges
- If β = 1, then the series inconclusive
Here [tex]a_{k}[/tex] = [tex]\frac{(x+2)^{k}}{\sqrt{k} }[/tex] and [tex]a_{k+1}[/tex] = [tex]\frac{(x+2)^{k+1}}{\sqrt{k+1} }[/tex]
Now limit is taken,
[tex]\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }|[/tex] = [tex]\lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|[/tex]
= [tex]\lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|[/tex]
= [tex]\lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|[/tex]
= [tex]|{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}[/tex]
= [tex]|{x+2 }|[/tex] < 1
- 1 < [tex]{x+2 }[/tex] < 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.
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