How do you determine the convergence or divergence of #sum_(n=1)^oo (-1)^n/((2n-1)!)#?

Answer 1

One can use the Ratio Test
#lim_(ntooo) |a_(n+1)/a_n|#

Given: #a_n = (-1)^n/((2n-1)!)#
#a_(n+1) = (-1)^(n+1)/((2n+1-1)!)#
#a_(n+1)/a_n = ((-1)^(n+1)/((2n+1-1)!))/((-1)^n/((2n-1)!))#
#a_(n+1)/a_n = ((-1)(-1)^(n)/((2n+1-1)!))/((-1)^n/((2n-1)!))#
#a_(n+1)/a_n = ((-1)/((2n+1-1)!))/((1)/((2n-1)!))#
#a_(n+1)/a_n = ((-1)/((2n)!))/((1)/((2n-1)!))#
#a_(n+1)/a_n = ((-1)/((2n)(2n-1)!))/((1)/((2n-1)!))#
#a_(n+1)/a_n = (-1(2n-1)!)/((2n)(2n-1)!)#
#a_(n+1)/a_n = -1/(2n)#
#lim_(ntooo) |a_(n+1)/a_n| = lim_(ntooo) 1/(2n) = 0#

This absolutely converges.

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Answer 2

You can determine the convergence or divergence of the series (\sum_{n=1}^\infty (-1)^n/((2n-1)!)) using the Alternating Series Test. This test states that if a series (\sum_{n=1}^\infty (-1)^n a_n) satisfies two conditions: 1) (a_{n+1} \leq a_n) for all (n) and 2) (\lim_{n \to \infty} a_n = 0), then the series converges.

For the given series, (a_n = \frac{1}{(2n-1)!}). Since (a_{n+1} = \frac{1}{(2(n+1)-1)!} = \frac{1}{(2n+1)!} \leq \frac{1}{(2n-1)!} = a_n) for all (n), the first condition is satisfied. Additionally, (\lim_{n \to \infty} \frac{1}{(2n-1)!} = 0). Therefore, both conditions are met, and by the Alternating Series Test, the series (\sum_{n=1}^\infty (-1)^n/((2n-1)!)) converges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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