The series \[\sum\limits_{{\text{m}} = 0}^\infty {\frac{1}{{{4^{\text{m}}}}}{{\left( {{\text{x}} – 1} \right)}^{2{\text{m}}}}} \] converges for A. -2 < X < 2 B. -1 < X < 3 C. -3 < X < 1 D. X < 3

[amp_mcq option1=”#NAME?” option2=”#NAME?” option3=”#NAME?” option4=”X < 3" correct="option1"]

The series $\sum\limits_{{\text{m}} = 0}^\infty {\frac{1}{{{4^{\text{m}}}}}{{\left( {{\text{x}} – 1} \right)}^{2{\text{m}}}}}$ is a geometric series with first term $\frac{1}{4}$ and common ratio $\left( {{\text{x}} – 1} \right)$. A geometric series converges if the absolute value of the common ratio is less than $1$. In this case, the absolute value of the common ratio is $\left| {{\text{x}} – 1} \right| < 1$, which means that $-1 < {\text{x}} < 3$. Therefore, the correct answer is $\boxed{{\text{B}}. -1 < {\text{x}} < 3}$.

Here is a brief explanation of each option:

  • Option A: $-2 < {\text{x}} < 2$. This is not the correct answer because the series diverges when $x=2$.
  • Option B: $-1 < {\text{x}} < 3$. This is the correct answer because the series converges when $-1 < x < 3$.
  • Option C: $-3 < {\text{x}} < 1$. This is not the correct answer because the series diverges when $x=1$.
  • Option D: $X < 3$. This is not the correct answer because the series converges when $-1 < x < 3$, and $X$ could be any value in this range.