Question

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.∑ₖ₌₂∞ ((x+3)ᵏ)/(k łn²k)

Accepted Answer

Identify the given power series: $$ \sum_{k=2}^{\infty} \frac{(x+3)^k}{k \ln^2(k)} . $$Notice that the series is centered at $$ x = -3 $$ because the term is $$ (x+3)^k . To $$find the radius of convergence, apply the Root Test or the Ratio Test. Here, the Root Test is convenient. Consider the general term $$ a_k = \frac{(x+3)^k}{k \ln^2(k)} . $$Compute $$ \lim_{k 	o \infty} \sqrt[k]{|a_k|} . $$Calculate $$ \sqrt[k]{|a_k|} = \sqrt[k]{\frac{|x+3|^k}{k \ln^2(k)}} = |x+3| \cdot \sqrt[k]{\frac{1}{k \ln^2(k)}} . As$$ $$ k 	o \infty $$, $$ \sqrt[k]{k} 	o 1 $$ and $$ \sqrt[k]{\ln^2(k)} 	o 1 $$, so the limit simplifies to $$ |x+3| . $$The Root Test states that the series converges if $$ \lim_{k 	o \infty} \sqrt[k]{|a_k|} \u003c 1 $$, so the radius of convergence $$ R $$ satisfies $$ |x+3| \u003c 1 . $$Thus, the radius of convergence is $$ R = 1 . $$Next, determine the interval of convergence by checking the endpoints $$ x = -3 - 1 = -4 $$ and $$ x = -3 + 1 = -2 . $$Substitute these values into the original series and analyze convergence using appropriate tests (such as the Comparison Test or Integral Test) because the behavior at endpoints depends on the series without the $$ (x+3)^k $$ factor.

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.

∑ₖ₌₂∞ ((x+3)ᵏ)/(k łn²k)

Verified video answer for a similar problem:

Key Concepts

Radius of Convergence

Interval of Convergence

Ratio Test for Convergence