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

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

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

Ratio for the absolute convergence is

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

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

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

So, when $x=-1+\mathrm{\pi }$

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

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

So, when $x=1+\pi$

$$\begin{gathered} \sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(x-\mathrm{\pi }\right)}^k=\sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(1+\mathrm{\pi }-\mathrm{\pi }\right)}^k \\ =\sum _{k=1}^{\infty }\frac{2k+1}{k^3}{\left(1\right)}^k \\ =\sum _{k=1}^{\infty }\frac{2k+1}{k^3} \end{gathered}$$
The result is just a constant multiple of the harmonic series which diverges

Therefore, the interval of convergence of the power series is$\left(-1+\mathrm{\pi },1+\mathrm{\pi }\right)$.