co but Inn- In(n - 1) = In In 1=0. n-1">
L.sufiya Khanam posted an Question
October 08, 2020 • 00:25 am 30 points
  • IIT JAM
  • Mathematics (MA)

Plz give me a detailed explanation on this 3 examples ,

(ii). V = (-1)": This sequence diverges whereas the sequence is bounded Si H - 1 V s1 Another example would be x = In n, since (iv). equ In n>co but Inn- In(n - 1) = In In 1=0. n-1 at it v). The sequence {sin(nt/2)),a diverges because the sequence is {1, 0, -1, 0, 1, 0, .), and hence it does not converge to any number.

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    Deepak singh 1 best-answer

    In first example - Vn=(-1)^n = {-1,1,-1,1...} limit point of Vn ={1,-1} (because 1 and -1 is coming infinitely times ) . Since limit not unique implies this series is not convergent. In second example - Xn = ln(n) Since ln(n) is unbounded so ,limit n tends to infinity Xn tends to infinity and thus divergent . On the other hand sequence Xn= ln(n/n-1) tends to 0 and hence convergent . In third example - An = {Sin(nπ/2)} limit of An ={ 1,0,-1,0,1,0......} (because 1,-1 and 0 appearing infinitely times ) Since limit not unique . hence not convergent. if any doubt then ask.

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    for 1st and 3rd I know it is convergent. how are they divergent as in 1st example it's like 1,-1,1,-1 .then it towards negative as well as positive . how it is divergent?

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    a sequence is convergent only if sequences have unique limit point..

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    but in 1 and 3 example , limit point is not unique , so not convergent

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    feel free to ask again

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    1 and 3 are oscillating right??

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    yes.

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    then in material it's given example for divergent sequence

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    a sequence is called divergent if limit is infinite or -infinite..( written in your book also )

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    but here in example 1 and 3 , limit is not infinite . so it is not divergent..it is oscillating

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    in your material , there is mistake..

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    feel free to ask again

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