Statement:
Let (an) be any sequence. And lim n -> infinity (an/an+1) = l if l > 1 then, (an) -> 0.
Proof:
Let, k be any real number such that 1 < k < l. Since, lim n ->infinity mod an/an+1 = l, there exist m belongs to N such that l - excellon < mod an/an+1 < l + excellon for all n >= m. Choosing excellon = l -k we obtain mod an/an+1 > k for all n >=m.
Now fix n >= m then, mod am/am+1 > k.
mod am+1/am+2 > k,.........mod an-1/an > k.
multiplying this inequalities we get,
mod am/am+1. mod am+1/am+2 ........mod an-1/an > k [k power n -m = k power n.k power -m
mod am/an > k power n -m. = k power n.1/m
Therefore, mod an/am < k power m (1/k) power n. = k power n.1/k power m
mod an < Ar power n. = k power m (1/k) power n]
Where, A = mod am k power m is a constant. r =1/k
Now. k > 1 => a < r < 1
Therefore, (r power n) -> 0.
Therefore, (an) -> 0.
Similarly, Let (an) be any (an) positive terms and lim n ->infinity an/an+1 = l. If l < 1, then (an) ->infinity.
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