Question

Find the interval of convergence for power series:$\sum _{k=0}^{\infty }\frac{k^2}{4^k+3}{\left(3x+7\right)}^k$

Short answer

The interval of convergence for power series is$\left(-\frac{8}{3},-2\right)$.

Step-by-step solution

Step 1. Given information. 

The given power series is$\sum _{k=0}^{\infty }\frac{k^2}{4^k+3}{\left(3x+7\right)}^k$.

Step 2. Find the interval of convergence.  

Let us assume $b_k=\frac{k^2}{4^k+3}{\left(3x+7\right)}^k$therefore $b_{k+1}=\frac{{\left(k+1\right)}^2}{4^{k+1}+3}{\left(3x+7\right)}^{k+1}$

Ratio for the absolute convergence is

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{{\left(k+1\right)}^2}{4^{k+1}+3}}{\left(3x+7\right)}^{k+1}}{{\displaystyle \frac{k^2}{4^k+3}}{\left(3x+7\right)}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|3x+7\right|{\left(\frac{k+1}{k}\right)}^2\frac{4^k+3}{4^{k+1}+3} \end{gathered}$$

Here the limit is $\left|3x+7\right|$So, by the ratio test of absolute convergence, we know that series will converge absolutely when$\left|3x+7\right|<1$that is$-\frac{8}{3}<x<-2$

Step 3. Find the interval of convergence.   

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

So, when $x=-\frac{8}{3}$

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{k^2}{4^k+3}{\left(3x+7\right)}^k=\sum _{k=0}^{\infty }\frac{k^2}{4^k+3}{\left(3\times -\frac{8}{3}+7\right)}^k \\ \sum _{k=0}^{\infty }\frac{k^2}{4^k+3}{\left(-1\right)}^k \end{gathered}$$

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

So, when$x=-2$

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{k^2}{4^k+3}{\left(3x+7\right)}^k=\sum _{k=0}^{\infty }\frac{k^2}{4^k+3}{\left(3\times -2+7\right)}^k \\ =\sum _{k=0}^{\infty }\frac{k^2}{4^k+3}{\left(1\right)}^k \\ \sum _{k=0}^{\infty }\frac{k^2}{4^k+3} \end{gathered}$$

The result is just a constant series which diverges.

Therefore, the interval of convergence of the power series is$\left(-\frac{8}{3},-2\right)$.