How do you find the interval of convergence #Sigma ((-1)^n(x+2)^n)/n# from #n=[1,oo)#?

Answer 1

The series:

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

is absolutely convergent for #x in (-3,-1)# and simply convergent for #x =-1#

To find the interval of convergence we can apply the ratio test, stating that a necessary condition for a series #sum_(n=1)^oo a_n# to converge is that:
#L = lim_(n->oo) abs (a_(n+1)/a_n) <= 1#
If #L < 1 # the condition is also sufficient and the series converges absolutely.

Let's calculate the ratio:

#abs (a_(n+1)/a_n) = abs (frac ( (x+2)^(n+1)/(n+1)) ((x+2)^n/n)) = abs(x+2) n/(n+1) #

so that:

#lim_(n->oo) abs (a_(n+1)/a_n) = abs(x+2)#
we can therefore conclude that the series is absolutely convergent for #abs (x+2) < 1# and divergent for #abs (x+2) > 1#.
In the case where #abs (x+2) = 1# we know that the test is inconclusive but we can see that for #x=-3# the series becomes:
#sum_(n=1)^oo (-1)^n(-1)^n/n = sum_(n=1)^oo 1/n#
that is divergent, while for #x= -1# we have:
#sum_(n=1)^oo (-1)^n/n#

that is convergent, but not absolutely convergent.

In conclusion the series converges for #x in (-3,-1]#
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Answer 2

To find the interval of convergence for the series (\sum_{n=1}^\infty \frac{(-1)^n(x+2)^n}{n}), we can use the Ratio Test.

Apply the Ratio Test:

  1. Compute the limit: (\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|).
  2. Determine whether the limit is less than 1, equal to 1, or greater than 1.

The Ratio Test states that if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If it's equal to 1, the test is inconclusive.

Applying the Ratio Test to the series (\sum_{n=1}^\infty \frac{(-1)^n(x+2)^n}{n}), we get:

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

For convergence, (|x + 2| < 1), thus the interval of convergence is (-3 < x < -1).

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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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