consider the power series\n∑(k = 1 to ∞) k^6(x + 4)^k/8^k·k^(20/3).\nfind the radius of convergence r.\nr…

Explanation:

Step1: Recall ratio - test formula

Let $a_k=\frac{k^{6}(x + 4)^{k}}{8^{k}\cdot k^{20/3}}$. Then $a_{k + 1}=\frac{(k+1)^{6}(x + 4)^{k+1}}{8^{k+1}\cdot(k + 1)^{20/3}}$. The ratio - test formula is $\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|$.

Step2: Calculate $\frac{a_{k + 1}}{a_{k}}$

[ \begin{align*} \frac{a_{k + 1}}{a_{k}}&=\frac{\frac{(k + 1)^{6}(x + 4)^{k+1}}{8^{k+1}\cdot(k + 1)^{20/3}}}{\frac{k^{6}(x + 4)^{k}}{8^{k}\cdot k^{20/3}}}\ &=\frac{(k + 1)^{6}(x + 4)^{k+1}}{8^{k+1}\cdot(k + 1)^{20/3}}\cdot\frac{8^{k}\cdot k^{20/3}}{k^{6}(x + 4)^{k}}\ &=\frac{(k + 1)^{6}}{k^{6}}\cdot\frac{k^{20/3}}{(k + 1)^{20/3}}\cdot\frac{x + 4}{8}\ &=\left(\frac{k + 1}{k}\right)^{6}\cdot\left(\frac{k}{k + 1}\right)^{20/3}\cdot\frac{x + 4}{8} \end{align*} ]

Step3: Find the limit as $k\rightarrow\infty$

We know that $\lim_{k\rightarrow\infty}\frac{k + 1}{k}=1$. So $\lim_{k\rightarrow\infty}\left(\frac{k + 1}{k}\right)^{6}=1$ and $\lim_{k\rightarrow\infty}\left(\frac{k}{k + 1}\right)^{20/3}=1$. Then $\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|=\left|\frac{x + 4}{8}\right|$.

Step4: Determine the radius of convergence

For convergence, $\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|\lt1$. So $\left|\frac{x + 4}{8}\right|\lt1$, which gives $|x+4|\lt8$. The radius of convergence $R = 8$.

Step5: Find the interval of convergence

We have the inequality $|x + 4|\lt8$, which is equivalent to $-8\lt x+4\lt8$. Subtracting 4 from all parts of the inequality, we get $-12\lt x\lt4$. Now we check the endpoints:

• When $x=-12$, the series becomes $\sum_{k = 1}^{\infty}\frac{k^{6}(-8)^{k}}{8^{k}\cdot k^{20/3}}=\sum_{k = 1}^{\infty}\frac{(- 1)^{k}}{k^{2/3}}$. This is an alternating - series. By the alternating - series test, since $\lim_{k\rightarrow\infty}\frac{1}{k^{2/3}} = 0$ and $\frac{1}{k^{2/3}}$ is a decreasing sequence, the series converges at $x=-12$.
• When $x = 4$, the series becomes $\sum_{k = 1}^{\infty}\frac{k^{6}(8)^{k}}{8^{k}\cdot k^{20/3}}=\sum_{k = 1}^{\infty}\frac{1}{k^{2/3}}$. This is a $p$ - series with $p=\frac{2}{3}\lt1$, so the series diverges at $x = 4$.

Answer:

$[-12,4)$