Q 40. - 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.

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

Step-by-Step Solution

Answer

The interval of convergence for power series is $\left[-\frac{3}{2},\frac{1}{2}\right]$.

Step 1. Given information.

The given power series is:

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

Step 2. Find the interval of convergence.

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

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

The ratio for the absolute convergence is 

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

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

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 $x=-\frac{3}{2}$

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

The result is just the alternating multiple of the harmonic series, which converges conditionally.

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

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

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

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