Proof:
Case (i):
Let, r =0. Then, (r power n) Reduces to the constant sequence. 0,0,0,.... and hence convergences to 0.
Case (ii):
Let, r =1. In this case (r power n) reduces to the constant sequence. 1,1,1,...... and hence convergences to 1.
Case(iii):
Let, r =-1. In this case (r power n) reduces to the -1,1,-1,1,.... which oscillate finiting the Oscillate Sequence.
Case(iv):
Let, r >1, then 0<1/r<1. And hence (1/r power n)->0. Therefore, (r power n) -> infinity.
Case(v):
Let, r <-1, then the terms of the sequence. r power n are alternatively posititve and negative. Also, mod r >1. And hence by, Case(iv) (mod r power n) is unbounded. Therefore, (r power n) Osciallates infinitely.
Case(vi):
Let, 0< r <1. In this case, (r power n) is a monotonnic decreasing sequence and (r power n) >0 for all n belongs to N. Therefore, (r power n) is a monotonnic decreasing and bounded below and hence (r power n) convergence. Let ( r power n) -> l. Since r power n>0 for all n,l >0------(1).
Claim:
l =0. Let excellon >0 be given. Since (r power n)->l, there exist m belongs to N there exist l < r power n< l+ excellon for all n>=m. Fix n>m, then l <r power n+1-----(2).
Also, r power n+1 =r.r power n < r( l +excellon)----(3).
Therefore, l < r ( l + excellon) by (2) and (3).
l < (r/1-r) excellon.
Therefore, this is true.
for every excellon >0,
we get, l <=0----(4).
Therefore, l = 0 (by (1)and (4))
Case(vii):
Let, -1 < r < 0.
Then, r power n=(-1) power n .mod r power n [when 0 <mod r < 1]
By case (vi) (mod r power n) ->0.
Therefore, Alse ((-1)power n) is bounded sequence.
Therefore, ((-1)power n.mod r power n) ->0.
Therefore, ( r power n) ->0.
Hence the statement.
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