question number 16. find the interval of convergence for ∑((-1)^k2^kx^k)/(5^(k + 1))

Explanation:

Step1: Apply ratio - test

Let $a_{k}=\frac{(-1)^{k}2^{k}x^{k}}{5^{k + 1}}$. Then $a_{k+1}=\frac{(-1)^{k + 1}2^{k+1}x^{k+1}}{5^{k+2}}$. Calculate the ratio $\left|\frac{a_{k + 1}}{a_{k}}\right|$. [ \begin{align*} \left|\frac{a_{k+1}}{a_{k}}\right|&=\left|\frac{\frac{(-1)^{k + 1}2^{k+1}x^{k+1}}{5^{k+2}}}{\frac{(-1)^{k}2^{k}x^{k}}{5^{k+1}}}\right|\ &=\left|\frac{(-1)^{k + 1}2^{k+1}x^{k+1}\cdot5^{k+1}}{(-1)^{k}2^{k}x^{k}\cdot5^{k+2}}\right|\ &=\left|\frac{(-1)\cdot2\cdot x}{5}\right|=\left|\frac{2x}{5}\right| \end{align*} ]

Step2: Find convergence condition

For convergence, $\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|<1$. So $\left|\frac{2x}{5}\right|<1$. Solve the inequality: [ \begin{align*} \left|\frac{2x}{5}\right|&<1\

• 1&<\frac{2x}{5}<1\ -\frac{5}{2}&<x<\frac{5}{2} \end{align*} ]

Step3: Check endpoints

Case 1: $x =-\frac{5}{2}$

The series becomes $\sum_{k = 0}^{\infty}\frac{(-1)^{k}2^{k}\left(-\frac{5}{2}\right)^{k}}{5^{k+1}}=\sum_{k = 0}^{\infty}\frac{(-1)^{k}2^{k}(-1)^{k}5^{k}}{2^{k}\cdot5^{k+1}}=\sum_{k = 0}^{\infty}\frac{1}{5}$. This is a divergent series (it is a constant - term series with non - zero constant).

Case 2: $x=\frac{5}{2}$

The series becomes $\sum_{k = 0}^{\infty}\frac{(-1)^{k}2^{k}\left(\frac{5}{2}\right)^{k}}{5^{k+1}}=\sum_{k = 0}^{\infty}\frac{(-1)^{k}2^{k}5^{k}}{2^{k}\cdot5^{k+1}}=\sum_{k = 0}^{\infty}\frac{(-1)^{k}}{5}$. This is a divergent series (by the divergence test, $\lim_{k\rightarrow\infty}\frac{(-1)^{k}}{5}$ does not exist).

Answer:

$\left(-\frac{5}{2},\frac{5}{2}\right)$