Find the interval of convergence for power series:$\sum _{k=0}^{\infty }\frac{k+1}{3^k}{\left(2x-5\right)}^k$

The given power series is$\sum _{k=0}^{\infty }\frac{k+1}{3^k}\left(2x-5\right)$.

Let us assume $b_k=\frac{k+1}{3^k}{\left(2x-5\right)}^k$therefore$b_{k+1}=\frac{\left(k+1\right)+1}{3^{k+1}}{\left(2x-5\right)}^{k+1}$

Ratio for the absolute convergence is

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{k+2}{3^{k+1}}}{(2x-5)}^{k+1}}{{\displaystyle \frac{k+1}{3^k}}{\left(2x-5\right)}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|2x-5\right|\frac{1}{3}\left(\frac{k+2}{k+1}\right) \end{gathered}$$

Here the limit is$\frac{1}{3}\left|2x-5\right|$ So, by the ratio test of absolute convergence, we know that series will converge absolutely when$\frac{1}{3}\left|2x-5\right|<1$that is$1<x<4$.

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

So, when $x=1$

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{k+1}{3^k}{\left(2x-5\right)}^k=\sum _{k=0}^{\infty }\frac{k+1}{3^k}{\left(2\times 1-5\right)}^k \\ =\sum _{k=0}^{\infty }\frac{k+1}{3^k}{\left(-3\right)}^k \\ =\sum _{k=0}^{\infty }k+1{\left(-1\right)}^k \end{gathered}$$

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

So, when $x=4$

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{k+1}{3^k}{\left(2x-5\right)}^k=\sum _{k=0}^{\infty }\frac{k+1}{3^k}{\left(2\times 4-5\right)}^k \\ =\sum _{k=0}^{\infty }\frac{k+1}{3^k}{\left(3\right)}^k \\ =\sum _{k=0}^{\infty }k+1 \end{gathered}$$

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

Therefore, the interval of convergence of the power series is$\left(1,4\right)$.