Replace x by StartFraction x Over 3 EndFraction minus 1 in the series ln (1 plus x )Superscript 5equals5 Summation from k equals 1 to infinity StartFraction (negative 1 )Superscript k plus 1 Baseline x Superscript k Over k EndFraction to obtain a power series for ln (StartFraction x Over 3 EndFraction )Superscript 5 centered at xequals3. What is the interval of convergence for the new power​ series? Choose the correct power series below. A. 5 Summation from k equals 1 to infinity StartFraction (negative 1 )Superscript k plus 1 Baseline (x minus 3 )Superscript k Over k EndFraction B. 5 Summation from k equals 1 to infinity StartFraction (negative 1 )Superscript k plus 1 Baseline (x minus 1 )Superscript k Over k EndFraction C. 5 Summation from k equals 1 to infinity StartFraction (negative 1 )Superscript k plus 1 Baseline (x minus 1 )Superscript k Over 3 Superscript k Baseline k EndFraction D. 5 Summation from k equals 1 to infinity StartFraction (negative 1 )Superscript k plus 1 Baseline (x minus 3 )Superscript k Over 3 Superscript k Baseline k EndFraction