Question

give an example of convergent sequence Xn of strictly positive numbers such that lim(Xn+1/Xn)=1
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Answer 1

Answer:

Yes. Suppose that there is some positive value c such that xi>ci for all i. Then, the average of the first n numbers exceeds c∑ni=1in=cn(n+1)2n=c(n+1)2 . This clearly diverges.

Thus, if the series does converge, the must not be such a constant, and therefore there is no lower bound above zero for xnn .

A similar proof assuming negative c such that xi<ci for all i proves there is no upper bound below zero and demonstrates the term in question converges to zero.

Answer 2

Given :   convergent sequence Xn of strictly positive numbers such that lim(Xn+1/Xn)=1

To Find : an example

Solution:

Xₙ  = 1/n   n belongs to positive integers

Xₙ₊₁  = 1/(n+1)

 Lim n→ ∞  | Xₙ₊₁ /Xₙ|

= Lim n→ ∞  | n/(n + 1)|

= Lim n→ ∞  | 1/(1 + 1/n)|

= 1/(1 + 0)

= 1/1

= 1

Hence Xₙ  = 1/n        is an example of convergent sequence Xn of strictly positive numbers such that lim(Xn+1/Xn)=1

Xₙ  = 1/n²   is another such example

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