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!}

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asked 2020-11-12
Radius and interval of convergence Determine the radius and interval of convergence of the following power series.
\(\sum_{k=0}^\infty\frac{(x-1)^k}{k!}\)

Expert Answers (1)

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
\((-\infty,\infty)\)
Answer:
The radius of convergence is
\(R=\infty\) Interval of convergence is given by:
\((-\infty,\infty)\)

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