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What is the value of the infinite sum #1+(x+1) + (x+1)^2+ (x+1)^3 + ....# Given #abs(x+1)<1# ?

Answer 1

Christopher Allers

Sum #= -1/x#

Here we have an infinite geometric progression with first term #a_1=1# and common ratio #r=(x+1)#

Since we are told that #abs(x+1)<1 -> abs(r)<1 # we know that the sum will converge to #a_1/(1-r)#

Thus, our infinite sum (S) in this example will converge to:

#S= 1/(1-(x+1)) = -1/x#

NB: Although this result may look strange at first sight, it is worth noticing that since #abs(x+1)<1 -># #-2< x<0#

In other words, #x# must be a negative number greater than #-2# for the sum to converge. This will make our result positive for all valid #x#

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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.

What is the value of the infinite sum #1+(x+1) + (x+1)^2+ (x+1)^3 + ....# Given #abs(x+1)&lt;1# ?

What is the value of the infinite sum #1+(x+1) + (x+1)^2+ (x+1)^3 + ....# Given #abs(x+1)<1# ?