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vodomira [7]
1 year ago
12

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

Mathematics
2 answers:
svlad2 [7]1 year ago
7 0

By the ratio test, the series converges for all x, since

\displaystyle \lim_{k\to\infty} \left|\frac{(k+1)^2 x^{k+4}}{(k+1)!} \cdot \frac{k!}{k^2 x^{k+3}}\right| = |x| \lim_{k\to\infty} \frac{(k+1)^2 k!}{k^2 (k+1)!} = |x| \lim_{k\to\infty} \frac1{k+1} = 0 < 1

Then the radius of convergence is R = ∞, and the interval of convergence is the entire real line, -∞ < x < ∞.

Hunter-Best [27]1 year ago
6 0

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

<h3>Find the radius of convergence R and the interval of convergence:</h3>

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 aₙ = (n²/n!) xⁿ⁺³  and  aₙ₊₁ = ((n+1)²/(n+1)!) xⁿ⁺¹⁺³ = ((n+1)²/(n+1)!) xⁿ⁺⁴

Now limit is taken,

\lim_{n \to \infty}  | \frac{a_{n+1} }{a_{n} } | = \lim_{n \to \infty}  | \frac{((n+1)^{2} /(n+1)!) x^{n+4}  }{(n^{2} /n!) x^{n+3}  } |

                      = \lim_{n \to \infty}  | \frac{((n+1)^{2} ) n!x^{n+4}  }{(n+1)!(n^{2} ) x^{n+3}  } |

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

                      =   \lim_{n \to \infty}  | \frac{ x  }{(n+1)} |  = 0 < 1

Since the limit is less than 1 the series is converging.

We get that,

interval of convergence = (-∞, ∞)

radius of convergence R = ∞

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

Learn more about power series here:

brainly.com/question/18763238

#SPJ1

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