Question

1. Find the radius and interval of convergence of the given series. ?_{n=0}^{?} frac{(x - 3)^{n+1}}{(n + 1)4^{n+1}} 2. Find the radius of convergence of the given series. ?_{n=1}^{?} [frac{2 · 4 · 6 · · · · · 2n}{3 · 5 · 7 · · · · · (2n + 1)}]x^{2n+1}

          1. Find the radius and interval of convergence of the given series.
?_{n=0}^{?} frac{(x - 3)^{n+1}}{(n + 1)4^{n+1}}

2. Find the radius of convergence of the given series.
?_{n=1}^{?} [frac{2 · 4 · 6 · · · · · 2n}{3 · 5 · 7 · · · · · (2n + 1)}]x^{2n+1}

$1. Find the radius and interval of convergence of the given series. ?n=0^? frac(x - 3)^n+1(n + 1)4^n+1 2. Find the radius of convergence of the given series. ?n=1^? [frac2 · 4 · 6 · · · · · 2n3 · 5 · 7 · · · · · (2n + 1)]x^2n+1$

Added by Cynthia S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Khushbu Rani

Step 1

For the first series, we can use the ratio test to find the radius of convergence: lim |(x-3)(n+2)(n+1)^(4n+5)/(x-3)(n+1)(n+2)^(4n+3)| = |x-3| lim |(n+1)/(n+2)|^(4n+5) = |x-3| lim [(n+1)/(n+2)]^4 * [(n+1)/(n+2)]^(1/2n+1) = |x-3| lim [(n+1)/(n+2)]^4 * 1 = |x-3| So Show more…

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Find the radius and interval of convergence of the given series: (x - 3)^(n+1) (n + 1)^(4n+1) n=0 2. Find the radius of convergence of the given series: 2.4.6^(2n) x^(en+1) 3.5^(7(2n + 1)) n=1