Answer to Question #116616 in Calculus for gly

Question #116616
find the radius and interval of convergence of given series

∑ (x+3)^n/2^n
n
=
0
1
Expert's answer
2020-05-20T18:57:52-0400

Given series is n=0un=n=0(x+3)n2n.\sum_{n=0}^{\infin} u_n = \sum_{n=0}^{\infin} \frac{(x+3)^n}{2^n}.

So, un=(x+3)n2nu_n = \frac{(x+3)^n}{2^n}

Now for convergence of series, limnun+1un<1\lim_{n\to \infin} |\frac{u_{n+1}}{u_n}| < 1

    limn(x+3)n+1/2n+1(x+3)n/2n<1\implies \lim_{n\to \infin} |\frac{(x+3)^{n+1}/2^{n+1}}{(x+3)^n/ 2^n}| < 1

    limnx+32<1    x+3<2\implies \lim_{n\to \infin} |\frac{x+3}{2}| < 1 \\ \implies |x+3| < 2

So, radius of convergence of given series is 2.


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