How do you show that the series #1/3-2/5+3/7-4/9+...+((-1)^(n-1)n)/(2n+1)+...# diverges?

Answer 1

By Divergence Test.

Please see the details below.

Let us consider:

#lim_(n to infty)n/(2n+1)#
By dividing the numerator and the denominator by #n#,
#=lim_(n to infty)1/(2+1/n)=1/(2+0)=1/2#,

which means that

#lim_(n to infty)((-1)^(n-1)n)/(2n+1)# does not exist. (#ne0#)

By Divergence Test, the given series diverges.

I hope that this was clear.

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Answer 2

To show that the series (\frac{1}{3} - \frac{2}{5} + \frac{3}{7} - \frac{4}{9} + \ldots + \frac{(-1)^{n-1}n}{2n+1} + \ldots) diverges, we can use the Alternating Series Test.

The Alternating Series Test states that if the terms of a series alternate in sign and decrease in absolute value, and the limit of the terms as (n) approaches infinity is zero, then the series converges. Conversely, if any of these conditions are not met, then the series diverges.

In this series, the terms alternate in sign and decrease in absolute value. However, the limit of the terms as (n) approaches infinity is not zero. To see this, consider the limit of the absolute value of the terms:

[\lim_{n \to \infty} \left| \frac{(-1)^{n-1}n}{2n+1} \right| = \lim_{n \to \infty} \frac{n}{2n+1} = \frac{1}{2}]

Since the limit of the absolute value of the terms is not zero, 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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