How do you determine the convergence or divergence of #Sigma ((-1)^n n!)/(1*3*5***(2n-1)# from #[1,oo)#?

Answer 1

The series converges.

We can apply d'Alembert's ratio test:

Suppose that;

# S=sum_(r=1)^oo a_n \ \ #, and #\ \ L=lim_(n rarr oo) |a_(n+1)/a_n| #

Then

if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

# S = sum_(n=0)^oo a_n # where #a_n = ( (-1)^n n! )/( 1.3.5 ... (2n-1) ) #

So our test limit is:

# L = lim_(n rarr oo) abs(( ( (-1)^(n+1) (n+1)! )/( 1.3.5 ... (2n-1).(2(n+1)-1) ) ) / ( ( (-1)^n n! )/( 1.3.5 ... (2n-1) ) )) # # \ \ \ = lim_(n rarr oo) abs(( (-1)^(n+1) (n+1)! )/( 1.3.5 ... (2n-1).(2(n+1)-1) ) * ( 1.3.5 ... (2n-1) ) / ( (-1)^n n! ) ) # # \ \ \ = lim_(n rarr oo) abs(( (-1) n!(n+1) )/( 2n+2-1 ) * ( 1 ) / ( n! )) # # \ \ \ = lim_(n rarr oo) abs(( (-1) (n+1) )/( 2n+1 )) # # \ \ \ = lim_(n rarr oo) abs(( n+1 )/( 2n+1 ) * (1/n)/(1/n)) # # \ \ \ = lim_(n rarr oo) abs(( 1+1/n )/( 2+1/n )) # # \ \ \ = 1/2 #

and we can conclude that the series converges

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Answer 2

To determine the convergence or divergence of the series (\sum_{n=1}^{\infty} \frac{(-1)^n n!}{1 \cdot 3 \cdot 5 \cdots (2n-1)}), we can use the ratio test.

The ratio test states that for a series (\sum_{n=1}^{\infty} a_n), if (\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L), then:

  • If (L < 1), the series converges absolutely.
  • If (L > 1) or the limit does not exist, the series diverges.
  • If (L = 1), the test is inconclusive.

For the given series, (a_n = \frac{(-1)^n n!}{1 \cdot 3 \cdot 5 \cdots (2n-1)}).

Let's find the ratio (R) of consecutive terms: [R = \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(-1)^{n+1} (n+1)!}{1 \cdot 3 \cdot 5 \cdots (2n+1)}}{\frac{(-1)^n n!}{1 \cdot 3 \cdot 5 \cdots (2n-1)}} \right|] [= \left| \frac{(-1)^{n+1} (n+1)! \cdot (2n-1)}{(-1)^n n! \cdot (2n+1)} \right|] [= \left| \frac{(n+1) \cdot (2n-1)}{2n+1} \right|]

Now, as (n) approaches infinity, (R) approaches (2), which is greater than (1). Therefore, by the ratio test, the series diverges.

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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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