$$\sum _n (-1)^n \frac {x^{2n+1}} {(2n+1)!}$$ I have solved this problem: the radius of convergence is coming to be infinity and hence the interval of convergence is $\Bbb R$. But in the answer it is given that the radius of convergence is $1$ and the interval of exact convergence is $[-1,1]$.
radius of convergence and exact interval of convergence
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sequences-and-series
power-series
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2YOU are right :-) D'Alembert rule for example... This is the development for $\sin x$. – 2017-01-12
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2But is my answer right ?I used d Alembert's rule here. – 2017-01-12
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2Your answer is correct. – 2017-01-12
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1Seems like somebody overlooked the $!$. The series $$\sum_n (-1)^n \frac{x^{2n+1}}{2n+1}$$ has radius of convergence $1$ and converges on $[-1,1]$. – 2017-01-12