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Determine whether the series is absolutely convergent 1-(1*3/3! (1*3*5/5!-(3*5*7)

User Dbjohn
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1 Answer

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It's not clear what your series is, so I'm going to take a wild guess on what it is you mean:


1-(1*3)/(3!)+(1*3*5)/(5!)-(1*3*5*7)/(7!)+\cdots

=\displaystyle\sum_(n=1)^\infty((-1)^(n-1))/((2n-1)!)\prod_(k=1)^n(2k-1)

For the sum to be absolute convergent, the sum of the absolute value of the summand must converge, so you are really examining the convergence of


\displaystyle\sum_(n=1)^\infty\frac1{(2n-1)!}\prod_(k=1)^n(2k-1)

This is easily checked with the ratio test:


\displaystyle\lim_(n\to\infty)\left|\frac{\displaystyle\frac1{(2(n+1)-1)!}\prod_(k=1)^(n+1)(2k-1)}{\displaystyle\frac1{(2n-1)!}\prod_(k=1)^n(2k-1)}\right|=\lim_(n\to\infty)\left|((1*3*5*\cdots*(2n-1)*(2n+1))/((2n+1)!))/((1*3*5*\cdots*(2n-1))/((2n-1)!))\right|

\displaystyle=\lim_(n\to\infty)\left|((2n+1)/((2n+1)(2n)))/(\frac11)\right|=\lim_(n\to\infty)\frac1{2n}=0<1

Since
\sum|a_n| converges by the ratio test, the series
\sum a_n converges absolutely.
User Kamagatos
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