91Ó°ÊÓ

The given power series is $\underset{k=1}{\overset{∞}{∑}}\frac{2k+1}{k^3}{\left(x-\mathrm{π}\right)}^k$.

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

Ratio for the absolute convergence is

$$\begin{gathered} \underset{k→∞}{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k→∞}{\lim }\left|\frac{{\displaystyle \frac{2(k+1)+1}{{\left(k+1\right)}^3}}{\left(x-\mathrm{π}\right)}^{k+1}}{{\displaystyle \frac{2k+1}{k^3}}{\left(x-\mathrm{π}\right)}^k}\right| \\ =\underset{k→∞}{\lim }\left|\frac{2k+3}{{\left(k+1\right)}^3}\left(x-\mathrm{π}\right)\frac{k^3}{2k+1}\right| \\ =\underset{k→∞}{\lim }\left|x-\mathrm{π}\right|{\left(\frac{k}{k+1}\right)}^3\frac{2k+3}{2k+1} \end{gathered}$$

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

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

So, when $x=-1+\mathrm{Ï€}$

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

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

So, when $x=1+\mathrm{Ï€}$

$$\begin{gathered} \underset{k=1}{\overset{∞}{∑}}\frac{2k+1}{k^3}{\left(x-\mathrm{π}\right)}^k=\underset{k=1}{\overset{∞}{∑}}\frac{2k+1}{k^3}{\left(1+\mathrm{π}-\mathrm{π}\right)}^k \\ =\underset{k=1}{\overset{∞}{∑}}\frac{2k+1}{k^3}{\left(1\right)}^k \\ =\underset{k=1}{\overset{∞}{∑}}\frac{2k+1}{k^3} \end{gathered}$$
The result is just a constant multiple of the harmonic series which diverges

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