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

Answer 1

#sum_ (n=1)^oo(-1)^(n+1)sechn# is convergent.

For alternating series, if the series is absolutely convergent, then it is also convergent. Now

#sech(n)=2/(e^n+e^(-n))# but
#1/(e^(n+1)+e^(-n-1)) < 1/(e^n+e^(-n))#

because

#1+e^(-2n) < e(1+e^(-2(n+1)))#
note that for #n > 0#
#1+e^(-2n) < 2# and #e = 2.71#
so #sech(n+1) < sech(n)# so
#sum_ (n=1)^oo(-1)^(n+1)sechn# is convergent.
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Answer 2

To determine the convergence or divergence of the series (\sum_{n=1}^{\infty} (-1)^{n+1}\text{sech}(n)) from (1) to (\infty), we use the alternating series test. This test states that if a series is alternating (i.e., terms alternate in sign) and the absolute value of the terms decreases as (n) increases, then the series converges.

In this case, (\text{sech}(n) = \frac{1}{\cosh(n)} = \frac{2}{e^n + e^{-n}}). Since (\cosh(n)) grows exponentially, (\text{sech}(n)) decreases exponentially as (n) increases.

Therefore, the series (\sum_{n=1}^{\infty} (-1)^{n+1}\text{sech}(n)) converges by 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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