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

\((-\infty,\infty)\)

Answer:

The radius of convergence is

\(R=\infty\) Interval of convergence is given by:

\((-\infty,\infty)\)