Q 31. - Calculus

Question

Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.

$\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{{\left(x-\frac{3}{2}\right)}^k}^{}$

Step-by-Step Solution

Answer

The interval of convergence for power series is $\left(\frac{1}{2},\frac{5}{2}\right)$.

Step 1. Given information.

The given power series is:

$\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{\left(x-\frac{3}{2}\right)}^k$

Step 2. Find the interval of convergence.

Let us assume $b_k={\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{\left(x-\frac{3}{2}\right)}^k$ therefore

$b_{k+1}={\left(-1\right)}^{k+1}\frac{\sqrt{k+1}}{{\left(k+1\right)}^3}{\left(x-\frac{3}{2}\right)}^{k+1}$

The ratio for the absolute convergence is 

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\left(-1\right)}^{k+1}{\displaystyle \frac{\sqrt{k+1}}{{\left(k+1\right)}^3}}{\left(x-{\displaystyle \frac{3}{2}}\right)}^{k+1}}{{\left(-1\right)}^k{\displaystyle \frac{\sqrt{k}}{k^3}}{\left(x-{\displaystyle \frac{3}{2}}\right)}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|\left(-1\right)\frac{k^3}{{\left(k+1\right)}^3}\frac{\sqrt{k+1}}{\sqrt{k}}\left(x-\frac{3}{2}\right)\right| \\ =\underset{k\to \infty }{\lim }\left|x-\frac{3}{2}\right|{\left(\frac{k}{k+1}\right)}^3\sqrt{\frac{k+1}{k}} \end{gathered}$$

Here the limit is $\left|x-\frac{3}{2}\right|$ So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\left|x-\frac{3}{2}\right|<1$ that is $\frac{1}{2}<x<\frac{5}{2}$.

Step 3. Find the interval of convergence.

Now, since the intervals are finite so we analyze the behavior of the series at the endpoints 

So, when $x=\frac{1}{2}$

$$\begin{gathered} \sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{\left(x-\frac{3}{2}\right)}^k=\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{\left(\frac{1}{2}-\frac{3}{2}\right)}^k \\ =\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{\left(-1\right)}^k \\ =\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^3}{\left(-1\right)}^{2k} \end{gathered}$$

The result is the alternating multiple of the harmonic series, which diverges.

So, when $x=\frac{5}{2}$

$$\begin{gathered} \sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{\left(x-\frac{3}{2}\right)}^k=\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{\left(\frac{5}{2}-\frac{3}{2}\right)}^k \\ =\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{\left(1\right)}^k \\ =\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^3}{\left(-1\right)}^k \end{gathered}$$
The result is just a constant multiple of the harmonic series which converges conditionally.

Therefore, the interval of convergence of the power series is $\sum _{k=1}^{\infty }{\left(-1\right)}^k\frac{\sqrt{k}}{k^3}{\left(x-\frac{3}{2}\right)}^k \mathrm{is} \left(\frac{1}{2},\frac{5}{2}\right)$.