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

Answer 1
Alternating series, which alternate between having positive and negative terms, often come in the forms #sum_(n=1)^oo(-1)^na_n# or #sum_(n=1)^oo(-1)^(n+1)a_n#. The only difference between these two is which terms are positive and which are negative.

Leibniz's rule, or the alternating series test, can be used to determine if one of these series converges or not.

For an alternating series such as #sum_(n=1)^oo(-1)^na_n# or #sum_(n=1)^oo(-1)^(n+1)a_n#, the sum will converge if both:
So, for #sum_(n=1)^oo(-1)^(n+1)/(2n-1)#, we see that the sequence being alternated is #a_n=1/(2n-1)#.
We see that #lim_(nrarroo)a_n=lim_(nrarroo)1/(2n-1)=lim_(nrarroo)(1/n)/(2-1/n)=0#. We also can see that as #nrarroo#, the denominator of #1/(2n-1)# will steadily grow, so #1/(2n-1)# will steadily decrease.
We can also show that #a_n>=a_(n+1)# by actually setting up that inequality:
#1/(2n-1)>=1/(2(n+1)-1)=>1/(2n-1)>=1/(2n+1)#
And then by cross-multiplying and solving this inequality, which is tedious, we can show that this is always true (at least for #n>1#).
Anyway, we've showed that #lim_(nrarroo)a_n=0# and #a_n>=a_(n+1)#, so we can conclude that #sum_(n=1)^oo(-1)^(n+1)/(2n-1)# converges.
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Answer 2

Arranging and adding successive terms

#1/(2k-1)-1/(2(k+1)-1)=1/(2k-1)-1/(2k+1)=2/((2k)^2-1)#

so

#sum_(k=1)^oo(-1)^(k+1)/(2k-1)=sum_(k=0)^oo 2/((2(2k+1))^2-1)#.

and we have

#sum_(k=0)^oo 2/((2(2k+1))^2-1) < sum_(k=0)^oo 2/(2(2k+1))^2<2sum_(k=1)^oo 1/k^2 = 2pi^2/6#

so

#sum_(k=1)^oo(-1)^(k+1)/(2k-1)# converges.
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Answer 3

To determine the convergence or divergence of the series ( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n-1} ) from ( n = 1 ) to ( \infty ), you can use the Alternating Series Test. This test states that if a series has alternating terms and meets two conditions: 1) the terms decrease in magnitude as ( n ) increases, and 2) the limit of the terms approaches zero as ( n ) approaches infinity, then the series converges. In this case, the terms ( \frac{(-1)^{n+1}}{2n-1} ) alternate in sign and decrease in magnitude as ( n ) increases. Additionally, the limit of the terms as ( n ) approaches infinity is zero. Therefore, by the Alternating Series Test, the series 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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