Question

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

Answer 1

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 < ∞.

Answer 2

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.  

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 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