Question
Download Solution PDFThe series \(\mathop \sum \limits_{{\rm{m}} = 0}^\infty \frac{1}{{{4^{\rm{m}}}}}{\left( {{\rm{x}} - 1} \right)^{2{\rm{m}}}}\) converges for
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDF\({{\rm{m}}_{{\rm{th}}}}{\rm{\;term}},{\rm{\;\;}}{{\rm{X}}_{\rm{m}}} = \frac{1}{{{4^{\rm{m}}}}}{\left( {{\rm{x}} - 1} \right)^{2{\rm{m}}}}\)
\({\left( {{\rm{m}} + 1} \right)^{{\rm{th}}}}{\rm{term}},{\rm{\;\;}}{{\rm{X}}_{{\rm{m}} + 1}} = \frac{1}{{{4^{{\rm{m}} + 1}}}}{\left( {{\rm{x}} - 1} \right)^{2{\rm{m}} + 2}}\)
\(\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {{\rm{m}} \to \infty } \end{array}\frac{{{{\rm{X}}_{{\rm{m}} + 1}}}}{{{{\rm{X}}_{\rm{m}}}}} = \frac{{{{\left( {{\rm{x}} - 1} \right)}^2}}}{4}\)
The series converges if \(\frac{{{{\left( {{\rm{x}} - 1} \right)}^2}}}{4} < 1\)
⇒ \( - 1 < {\rm{x}} < 3\).
Last updated on May 28, 2025
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