Find the interval of convergence of the series.
-$$\sum _ { n = 0 } ^ { \infty } \frac { ( x - 3 ) ^ { n } } { 4 n + 4 }$$
A) $2 \leq x

Question

Find the interval of convergence of the series.
-$$\sum _ { n = 0 } ^ { \infty } \frac { ( x - 3 ) ^ { n } } { 4 n + 4 }$$
A) $2 \leq x < 4$
B) $x < 4$
C) $2 < x < 4$
D) $- 1 \leq x < 7$

Quizplus · Accepted Answer

The interval of convergence can be found using the ratio test, which involves taking the limit of the absolute value of the ratio of consecutive terms. For the given series, the terms are $\frac{(x-3)^n}{4n+4}$. Applying the ratio test:$$\lim _ { n \rightarrow \infty } \left| \frac { ( x - 3 ) ^ { n + 1 } / ( 4 n + 8 ) } { ( x - 3 ) ^ { n } / ( 4 n + 4 ) } \right| = \lim _ { n \rightarrow \infty } \left| \frac { ( x - 3 ) ( 4 n + 4 ) } { 4 n + 8 } \right| = | x - 3 | \lim _ { n \rightarrow \infty } \frac { 4 n + 4 } { 4 n + 8 } = | x - 3 |$$For the series to converge, this limit must be less than 1:$$| x - 3 | < 1$$Solving this inequality:$$-1 < x - 3 < 1$$$$2 < x < 4$$However, we must also check the endpoints $x = 2$ and $x = 4$ to determine if they are included in the interval of convergence. For $x = 2$, the series becomes $\sum _ { n = 0 } ^ { \infty } \frac { ( -1 ) ^ { n } } { 4 n + 4 }$, which converges by the Alternating Series Test.For $x = 4$, the series becomes $\sum _ { n = 0 } ^ { \infty } \frac { 1 } { 4 n + 4 }$, which diverges because it is essentially a harmonic series.Therefore, the interval of convergence is $2 \leq x < 4$.

Find the Interval of Convergence of the Series ∑n=0∞(x−3)n4n+4\sum _ { n = 0 } ^ { \infty } \frac { ( x - 3 ) ^ { n } } { 4 n + 4 }n=0∑∞​4n+4(x−3)n​

Find the Interval of Convergence of the Series $\sum _ { n = 0 } ^ { \infty } \frac { ( x - 3 ) ^ { n } } { 4 n + 4 }$