What is the interval of convergence of #sum_1^oo (-2)^n(n+1)(x-1)^n #?

Answer 1

#(1/2,3/2]#

Apply the test of ratios.

The infinite sum #sum^ooa_n# converges when
#lim_(nrarroo)abs((a_(n+1))/(a_n))<1#

This provides

#lim_(nrarroo)abs(((-2)^(n+1)(n+2)(x-1)^(n+1))/((-2)^n(n+1)(x-1)^n))<1#

Condensed:

#lim_(nrarroo)abs((-2(n+1)(x-1))/(n+2))<1#

Assessing the maximum yields

#abs(-2(x-1))<1#

leading to the disparity

#1/2 < x < 3/2#
We now have to plug in #x=1/2# and #x=3/2# to see if the sum converges at the endpoints.
Plugging in #x=1/2# does not converge, but #x=3/2# does.

Thus, the response is

#1/2 < x <=3/2# or #(1/2,3/2]#
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Answer 2

The interval of convergence of the series ∑ (-2)^n(n+1)(x-1)^n is (-1, 3), inclusive of -1 and exclusive of 3.

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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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