Q 35. - 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.

$\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(x+2\right)}^k$

Step-by-Step Solution

Answer

The interval of convergence for power series is $\left(-4,0\right)$.

Step 1. Given information.

The given power series is:

$\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(x+2\right)}^k$

Step 2. Find the interval of convergence.

Let us assume $b_k=\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(x+2\right)}^k$ therefore

$b_{k+1}=\frac{{\left[\left(k+1\right)!\right]}^2}{\left[2\left(k+1\right)\right]!}{\left(x+2\right)}^{k+1}$

The ratio for the absolute convergence is 

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

Here the limit $\frac{1}{2}\left|x+2\right|$ is So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\frac{1}{2}\left|x+2\right|<1$ that is $-4<x<0$.

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 $x=-4$

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(x+2\right)}^k=\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(-4+2\right)}^k \\ =\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(-2\right)}^k \end{gathered}$$

The result is just a constant multiple of the arithmetic series, which diverges conditionally.

So, when $x=0$

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(x+2\right)}^k=\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(0+2\right)}^k \\ =\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(2\right)}^k \end{gathered}$$

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

Therefore, the interval of convergence of the power series $\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^2}{\left(2k!\right)}{\left(x+2\right)}^k$ is $\left(-4,0\right)$.