What is the interval of convergence of $sum_1^oo [(-x)^(2n+1)]/[(2n+1)!]$ ?

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Answer 1 · 2017-01-05T07:34:22

The interval of convergence is $x in ] -oo,+oo [$

We use the ratio test

If the $lim_(n->oo)∣a_(n+1)/a_n∣=L$

If $L<1$ , then $sum_1 ^oa_n$ converges

If $L>1$ , then $sum_1 ^oa_n$ diverges

If $L=1$ , the test is inconclusive

$(∣a_(n+1)/a_n∣) =(∣((-x)^(2(n+1)+1)/((2(n+1)+1)!))/((-x)^(2n+1)/((2n+1)!))∣)$

Now, we calculate the limits

$lim_(n->oo)(∣((-x)^(2(n+1)+1)/((2(n+1)+1)!))/((-x)^(2n+1)/((2n+1)!))∣)$

$=lim_(n->oo)(∣((-x)^(2n+3)/((2n+3)!))/((-x)^(2n+1)/((2n+1)!))∣)$

$=lim_(n->oo)(∣((-x)^(2n+3)/((-x)^(2n+1))*(((2n+1)!)/((2n+3)!))∣)$

$=lim_(n->oo)(∣x^2*(1/((2n+3)(2n+2))∣)$

$=lim_(n->oo)(∣x^2*(1/((2n+3)2(n+1))∣)$

$=∣(x^2/2)∣*lim_(n->oo)(∣1/((2n+3)(n+1))∣)$

$=∣(x^2/2)∣*0$

$=0$

Therefore,

$L<1$ , for every x,

So $sum_1 ^oox^(2n+1)/((2n+1)!)$ converges for all x

The interval is $-oo < x <+oo$

What is the interval of convergence of sum_1^oo [(-x)^(2n+1)]/[(2n+1)!]sum_1^oo [(-x)^(2n+1)]/[(2n+1)!]?