Q 34. - 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.

$\sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{x}^k$

Step-by-Step Solution

Answer

The interval of convergence for power series is $\sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{x}^k$.

Step 1. Given information.

The given power series is:

$\sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{x}^k$

Step 2. Find the interval of convergence.

Let us assume $b_k= \frac{1}{3^k+5^k}{x}^k$ therefore

$b_{k+1}= \frac{1}{3^{k+1}+5^{k+1}}{x}^{k+1}$

The ratio for the absolute convergence is 

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{1}{3^{k+1}+5^{k+1}}{x}^{k+1}}}{{\displaystyle \frac{1}{3^k+5^k}{x}^k}}\right| \\ =\underset{k\to \infty }{\lim }\left|\frac{3^k+5^k}{3^{k+1}+5^{k+1}}x\right| \\ =\underset{k\to \infty }{\lim }\left|x\right|\frac{3^k+5^k}{3^{k+1}+5^{k+1}} \end{gathered}$$

Here the limit $\left|x\right|$ is 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 $x=-1$

$$\begin{gathered} \sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{x}^k=\sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{\left(-1\right)}^k \\ =\sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{\left(-1\right)}^k \end{gathered}$$

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

So, when $x=1$

$$\begin{gathered} \sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{x}^k=\sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{\left(1\right)}^k \\ \sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{\left(1\right)}^k \end{gathered}$$

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

Therefore, the interval of convergence of the power series $\sum _{k=0}^{\infty } \frac{1}{3^k+5^k}{x}^k$ is $\left(-1,1\right)$.