Answer
The interval of convergence for the series is: $(1,3)$ and the radius of convergence is $R=1$.
Now, we have the sum: $\Sigma_{k=0}^{\infty} (x-2)^{k}=\dfrac{1}{3-x} $
Work Step by Step
Ratio Test: Let us consider a series $\Sigma a_k$ whose limit $l$ can be obtained as: $ l=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|$
1. For $l \lt 1$, the series is absolutely convergent.
2. For $l \gt 1$, the series is divergent.
3. For $l = 1$, the series is inconclusive.
Therefore, $ L=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty} \dfrac{(x-2)^{k+1}}{(x-2)^{k}} \\=\lim\limits_{k \to \infty} |x-2|$
So, we can conclude that the given series converges absolutely for $|x-2| \lt 1$ and diverges for $|x-2| \gt 1$ by the ratio test. But the test fails for the value of $x=1$ and $x=3$.
$\bf{Case -1}:$ For $x=1$, the given series becomes: $\Sigma_{k=0}^{\infty} (1-2)^{k}= \Sigma_{k=0}^{\infty}(-1)^k$; a divergent series.
$\bf{Case -2}:$ For $x=3$, the given series becomes: $\Sigma_{k=0}^{\infty} (3-2)^{k}= \Sigma_{k=0}^{\infty}(1)$; a divergent series.
Thus, the interval of convergence for the series is: $(1,3)$ and the radius of convergence is $R=1$.
Now, we have the sum: $\Sigma_{k=0}^{\infty} (x-2)^{k}= \dfrac{1}{1-(x-2)}=\dfrac{1}{3-x} $
