Find the interval of convergence for power series:$\sum _{k=1}^{\infty }\frac{1}{k^2}{\left(x+3\right)}^k$

The given power series is$\sum _{k=1}^{\infty }\frac{1}{k^2}{\left(x+3\right)}^k$.

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

Ratio for the absolute convergence is

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

Here the limit is$\left|x+3\right|$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 analyse the behavior of the series at the endpoints.

So, when $x=-4$

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

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

Also, when $x=-2$

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

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

Therefore, the interval of convergence of the power series is$[-4,-2)$