91Ó°ÊÓ

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

$b_{k+1}=\frac{1}{3^{k+1}+5^{k+1}}{\left(x+3\right)}^{k+1}$

The ratio for the absolute convergence is

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

Here the limit $\left|x+3\right|$is So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\left|x+3\right|<1$that is$-4<x<-2$.

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

So, when $x=-4$

$$\begin{gathered} \underset{k=0}{\overset{∞}{∑}}\frac{1}{3^k+5^k}{\left(x+3\right)}^k=\underset{k=0}{\overset{∞}{∑}}\frac{1}{3^k+5^k}{\left(-4+3\right)}^k \\ =\underset{k=0}{\overset{∞}{∑}}\frac{1}{3^k+5^k}{\left(-1\right)}^k \end{gathered}$$

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

So, when $x=-2$

$$\begin{gathered} \underset{k=0}{\overset{∞}{∑}}\frac{1}{3^k+5^k}{\left(x+3\right)}^k=\underset{k=0}{\overset{∞}{∑}}\frac{1}{3^k+5^k}{\left(-2+3\right)}^k \\ =\underset{k=0}{\overset{∞}{∑}}\frac{1}{3^k+5^k}{\left(1\right)}^k \end{gathered}$$

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

Therefore, the interval of convergence of the power series $\underset{k=0}{\overset{∞}{∑}}\frac{1}{3^k+5^k}{\left(x+3\right)}^k$is $\left(-4,-2\right)$.