Radius of Convergence and Interval of Convergence

Question

The questions ask to find the ROC and IOC of: $$\frac{1}{k^{1.001}}(5^{0.5}x)^k$$

Answer 1

Both the ratio and root test are applicable for this series. Rewrite $$\sum\limits_{k =1}^{\infty}\dfrac{1}{k^{1.001}}(5^{0.5}x)^k = \sum\limits_{k =1}^{\infty}\dfrac{(\sqrt{5})^k}{k^{1.001}}x^k = \sum\limits_{k =1}^{\infty}a_k x^k$$ with $a_k=\dfrac{(\sqrt{5})^k}{k^{1.001}}.$ Then radius of convergence is $$ {R}={\lim\limits_{k \to\infty} {\dfrac{a_k}{a_{k+1}}}} = \lim\limits_{k \to\infty} {\dfrac{(\sqrt{5})^k}{k^{1.001}}\cdot\dfrac{(k+1)^{1.001}}{(\sqrt{5})^{k+1}}}=\dfrac{1}{\sqrt{5}},$$ since $\lim\limits_{k \to\infty}{\dfrac{(k+1)^{1.001}}{k^{1.001}}}=\lim\limits_{k \to\infty}{\left(\dfrac{k+1}{k}\right)^{1.001}}=\left(\lim\limits_{k \to\infty}{\dfrac{k+1}{k}}\right)^{1.001}=1.$