Question # Radius and interval of convergence Determine the radius and interval of convergence of the following power series. sum_{k=0}^inftyfrac{(x-1)^k}{k!}

Series
ANSWERED Radius and interval of convergence Determine the radius and interval of convergence of the following power series.
$$\sum_{k=0}^\infty\frac{(x-1)^k}{k!}$$ 2020-11-13

Given: Series
$$\sum_{k=0}^\infty\frac{(x-1)^k}{k!}$$
To Find: Radius and Interval of Convergence of given series
Concept Used: Ratio Test:
For a series with terms $$\left\{a_n\right\}$$, consider the ratio
$$\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=L$$
If $$L < 1$$ then the series converges
Calculations:
Consider the ratio of the given series as follows:
$$\left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{(x-1)^{k+1}}{(k+1)!}\frac{k!}{(x-1)^k}\right|$$
$$\Rightarrow\left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{(x-1)^{k+1-k}k!}{(k+1)k!}\right|$$
$$\Rightarrow\left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{x-1}{k+1}\right|=\frac{1}{k+1}|x-1|$$
$$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\frac{|x-1|}{k+1}=|x-1|\lim_{k\to\infty}\frac{1}{k+1}$$
$$\Rightarrow\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=|x-1|(0)=0$$
Hence, $$L = 0$$ for all values of x and so the given series converges for all x
The radius of convergence is given by :
$$R=\frac{1}{L}=\frac10=\infty$$
And hence, the interval of convergence is
$$(-\infty,\infty)$$
$$R=\infty$$ Interval of convergence is given by:
$$(-\infty,\infty)$$