Q 32. - 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.

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

Step-by-Step Solution

Answer

The interval of convergence for power series is $\left(-1,1\right)$.

Step 1. Given information.

The given power series is:

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

Step 2. Find the interval of convergence.

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

$b_{k+1}=\frac{2^{k+1}+1}{\sqrt{k+1}}{x}^{k+1}$

The ratio for the absolute convergence is 

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

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

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 $x=-1$

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

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

So, when $x=1$

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

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

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