Given series is ∑n=0∞un=∑n=0∞(x+3)n2n.\sum_{n=0}^{\infin} u_n = \sum_{n=0}^{\infin} \frac{(x+3)^n}{2^n}.∑n=0∞un=∑n=0∞2n(x+3)n.
So, un=(x+3)n2nu_n = \frac{(x+3)^n}{2^n}un=2n(x+3)n
Now for convergence of series, limn→∞∣un+1un∣<1\lim_{n\to \infin} |\frac{u_{n+1}}{u_n}| < 1limn→∞∣unun+1∣<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⟹limn→∞∣(x+3)n/2n(x+3)n+1/2n+1∣<1
⟹ limn→∞∣x+32∣<1 ⟹ ∣x+3∣<2\implies \lim_{n\to \infin} |\frac{x+3}{2}| < 1 \\ \implies |x+3| < 2⟹limn→∞∣2x+3∣<1⟹∣x+3∣<2
So, radius of convergence of given series is 2.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment