How do you determine if the series the converges conditionally, absolutely or diverges given #Sigma (-1)^(n+1)/(n+1)^2# from #[1,oo)#?

Answer 1

See below.

Considering that

#L=sum_(k=1)^oo 1/k^2 = pi^2/6# (Basel problem)

we have

#L = sum_(k=1)^oo 1/(2k)^2 + sum_(k=1)^oo 1/(2k-1)^2 = S_p+S_i#

but

#S_p = 1/4 L# then
#S_p-S_i = 1/4L-(L-1/4L)= -L/2 = -pi^2/12#

then

#sum_(k=1)^oo(-1)^k/k^2 = -pi^2/12# so converges
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Answer 2

#sum_(n=1)^oo (-1)^(n+1)/(n+1)^2#

is absolutely convergent.

By direct comparison we can see that for #n>=1#:
#1/(n+1)^2 < 1/n^2#

As:

#sum_(n=1)^oo 1/n^2#
is convergent based on the #p#-series test, then also:
#sum_(n=1)^oo 1/(n+1)^2#

is convergent, and:

#sum_(n=1)^oo (-1)^(n+1)/(n+1)^2#

is absolutely convergent.

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Answer 3

To determine the convergence behavior of the series ( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)^2} ), we can use the alternating series test and the absolute convergence test.

First, let's check for absolute convergence:

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

This series is a convergent p-series with ( p = 2 ), so it converges absolutely.

Next, let's check for conditional convergence using the alternating series test. The series ( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)^2} ) alternates signs and the absolute values of its terms decrease monotonically to zero.

Therefore, since the series converges absolutely, it also converges conditionally.

In summary:

  • The series converges absolutely because ( \sum_{n=1}^{\infty} \frac{1}{(n+1)^2} ) converges.
  • The series also converges conditionally because it satisfies the conditions of the alternating series test.
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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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