Find the interval of convergence of the power series, where c &gt; 0 and k is a positive integer. ∑n=1∞n!(x−c)n1⋅3⋅5⋅…⋅(2n−1)

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Find the interval of convergence of the power series, where c &amp;gt; 0 and k is a positive integer.  ∑n=1∞n!(x−c)n1⋅3⋅5⋅…⋅(2n−1)

wheezym · Accepted Answer

To determine: The interval of the convergence of the power series, where c &amp;gt; 0 and k is a positive integer. ∑n=1∞n!(x−c)n1⋅3⋅5⋅…⋅(2n−1) Calculation: Let ∑n=1∞an(x−c)n=∑n=1∞n!(x−c)n1⋅3⋅5⋅…⋅(2n−1) By comparing an=n!1⋅3⋅5⋅…⋅(2n−1) By using the ratio test, L=limn→∞|an+1an|=|(n+1)!(x−c)n+11⋅3⋅5⋅…⋅(2n+1)×1⋅3⋅5⋅…⋅(2n−1)n!| L=limn→∞|(n+1)(2n−1)(x−c)(2n+1)| L=|x−c|limn→∞|(n+1)(2n−1)(2n+1)| But limn→∞|(n+1)(2n−1)2n+1|=∞ So, the radius of convergence is 0 and the power series converges at x=c ⇒|x−c|&amp;lt;0 ⇒−c&amp;lt;x&amp;lt;c But c &amp;gt;0 So, it is 0 Therefore, the interval of the convergence for the given power series is (0,c]

Find the interval of convergence of the power series, where c > 0 and k is a positive integer. sum_{n=1}^inftyfrac{n!(x-c)^n}{1cdot3cdot5cdot...cdot(2n-1)}

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